CHAPTER 13 Chemical Kinetics: Clearing the Air

CHAPTER 13 Chemical Kinetics: Clearing the Air 13.1. Collect and Organize For the plot of Figure P13.1, we are to identify which curves represent [N O] and [O ] over time for the conversion of N O to N and O according to the equation N O(g) N (g) + O (g) As the reaction proceeds, the concentration of the reactant, N O, decreases and the concentration of the product, O, increases. The rate at which N O is used up in the reaction is twice the rate at which O is produced. The green line represents [N O], and the red line represents [O ]. [N ], represented by the blue line, increases twice as fast as [O ] because two N molecules are produced for every one O molecule in the reaction. 13.. Collect and Organize For the plot of Figure P13., we are to identify which curves represent [SO ] and [O ] over time for the reaction of SO with O to form SO 3 according to the equation 1 SO ( g) + O ( g) SO ( g ) 3 As the reaction proceeds, the concentrations of the reactants, SO and O, decrease over time. The rate at which SO decreases in the reaction is twice the rate at which O is used up according to the balanced equation. The blue line represents [SO ], and the green line represents [O ]. Because one molecule of SO 3 is produced for every molecule of SO reacted, [SO 3 ] (red line) increases at the same rate at which [SO ] decreases. 13.3. Collect and Organize For three different initial concentrations of reactant A shown in Figure P13.3, we are to choose which would have the fastest rate for the conversion A B. We are given that the reaction is second order in A. The rate law is written as follows: Rate = k[a] As the concentration of A increases, the rate increases. Figure P13.3(b) has the fastest reaction rate because it has the highest concentration of A. The higher the concentration of reactant molecules, the more often they collide, which increases the rate of the reaction. 39

40 Chapter 13 13.4. Collect and Organize Knowing that the rate of the reaction A + B C is first order in both A and B, we are to choose the sample in Figure P13.4 that produces the fastest initial rate. The initial rate of a reaction depends on the concentration of the reactants involved in the rate law epression. For the reaction described in this problem, the rate law would be Rate = k[a][b] Therefore, the higher the concentrations of both A and B, the faster the rate of the reaction. Figure P13.4(c) has the highest concentration of A (seven red spheres) and B (si blue spheres) compared to sample (a), with 4 A + 5 B, and sample (b), with 6 A + 5 B, so sample (c) has the fastest initial rate. The lower the concentration of the reactants, the slower the initial rate of the reaction. In this problem, sample (a) would give the slowest initial rate. 13.5. Collect and Organize For the reaction profile in Figure P13.5, we are to identify the parts of the energy diagram. The energy of the reactants is indicated at the beginning of the reaction; the energy of the products is indicated at the end of the reaction. The activation energy is the energy that the reaction must attain to form products; for the forward reaction it will be the energy change from the reactants to the energy of the transition state, and for the reverse reaction it will be the energy change from the products to the energy of the transition state. The energy change of the reaction is the energy difference between that of the products and that of the reactants. (a) The energy of the reactants is shown as (1). (b) The energy of the products is shown as (5). (c) The activation energy of the forward reaction is shown as (). (d) The activation energy for the reverse reaction is shown as (4). (e) The energy change of the reaction is shown as (3). Be careful not to assume that reaction (b) is slow because it is nonspontaneous. The rate of a reaction does not depend on the thermodynamics of the reactants and the products. 13.6. Collect and Organize From the five reaction profiles in Figure P13.6, we are to choose the ones that corresponds in overall reaction energy and magnitude of activation energy and stability of an intermediate.

Chemical Kinetics 41 A reaction profile for an endothermic reaction will show the energy of the products higher than the energy of the reactants; a reaction profile for an eothermic reaction will show the energy of the products lower than the energy of the reactants. The E a is the energy difference between the reactants (on the left-hand side of the graph) and the transition state (the highest point on the reaction profile curve). Large activation energies are shown by a large difference between the transition-state energy and the reactant energy; small activation energies are shown by a small difference between the transition-state energy and the reactant energy. A reaction with a stable intermediate will show a low-energy dip in the middle of the reaction profile. (a) Reaction profile 3 shows a highly eothermic reaction with a large activation energy. (b) Reaction profile shows a highly endothermic reaction with a moderate activation energy. (c) Reaction profile 5 shows a reaction that forms a stable intermediate. Reaction profile 1 shows an eothermic reaction with a small activation energy, and reaction profile 4 shows a reaction profile that has little or no enthalpy change but also shows a barrier over which the reactants must proceed to form products. 13.7. Collect and Organize For the three reaction profiles in Figure P13.7, we are to choose the one that has the smallest rate constant.

4 Chapter 13 The rate of a reaction (and therefore the rate constant) is determined by the activation energy (E a ) of the slowest step. All the reactions shown consist of a single step. The E a is the energy difference between the reactants (on the left-hand side of the graph) and the transition state (the highest point on the reaction profile curve). Reaction profile (b) has the largest E a and therefore is the slowest reaction. Be careful not to assume that reaction (b) is slow because it is nonspontaneous. The rate of a reaction does not depend on the thermodynamics of the reactants and the products. 13.8. Collect and Organize We are given three reaction profiles in Figure P13.8 and, from the relative activation energies (E a ), we can determine which reaction has the largest rate constant. The activation energy is the energy barrier that the reactants must overcome to form products; the lower the activation energy barrier, the faster the reaction and the greater the rate constant for the reaction. Reaction profile (a) has the lowest E a (energy difference from reactants to the transition state at the highest point in energy in the reaction profile) and therefore has the fastest reaction. Profile (c) represents the reaction with the slowest rate. The ΔE for this reaction is zero. 13.9. Collect and Organize We are to match the reaction profile in Figure P13.9 with the possible reactions given. The reaction profile shows a two-step reaction that has a slightly larger activation energy for its second step than for its first step. Reaction (c) is correct because it is a two-step reaction with the first step faster than the second. Reaction (b), which occurs in a single step, would show only one transition state and one activation energy in its reaction profile. 13.10. Collect and Organize We are to choose the mechanism consistent with the reaction profile in Figure P13.10 showing two activation energies.

Chemical Kinetics 43 The presence of two activation energies in the reaction profile means that two steps are in the mechanism. The relative activation energies of the two steps tell us the relative rates of the two steps, with the step having the smaller E a being the fastest. Reaction mechanism (a) is consistent with the reaction profile because two transition states are shown and the activation energy of the second step is lower than that of the first step, so the second step proceeds at a faster rate than the first step. A single-step reaction, as in (b), would give a reaction profile showing only a single transition state and one activation energy barrier. 13.11. Collect and Organize Given the reaction profile of an uncatalyzed reaction (Figure P13.11), we are to choose the reaction profile corresponding to the catalyzed reaction. A catalyst increases the rate of reaction by decreasing the activation energy of the reaction through an alternate pathway to the products. This alternate pathway usually involves more steps. Reaction profile (b) correctly shows the catalyzed reaction. Reaction profiles (a) and (c) cannot be correct because the initial uncatalyzed nonspontaneous reaction is represented as spontaneous. A catalyst cannot change a nonspontaneous reaction into a spontaneous reaction. 13.1. Collect and Organize Given the reaction profile of an uncatalyzed reaction (Figure P13.1), we are asked to choose the reaction profile for the catalyzed reaction. A catalyst increases the rate of a reaction by lowering the activation energy barrier by providing a different, lower-energy path from reactants to products. Figure P13.1(c) best represents the reaction profile of the catalyzed reaction because the largest activation energy barrier of the catalyzed reaction is lower than the energy barrier of the uncatalyzed reaction. The difference in pathways is shown by the two transition states present in the catalyzed reaction profile compared to the one transition state shown in the uncatalyzed reaction profile.

44 Chapter 13 Reaction profiles (a) and (b) cannot be correct because in both the relative energy of the products and reactants is different from the uncatalyzed reaction; the energy of the reactants and the energy of the products do not change in a catalyzed reaction. 13.13. Collect and Organize Of the highlighted elements in Figure P13.13, we are to choose which forms gaseous oides associated with photochemical smog. Photochemical smog is the result of sunlight interacting with NO produced by automobile emissions and volatile organic compounds (VOCs) released into the atmosphere. Nitrogen (light blue) forms the volatile oides that are components of photochemical smog. Sunlight causes a reaction of NO and VOCs to produce peroyacyl nitrates that are very irritating to the lungs. 13.14. Collect and Organize Of the highlighted elements in Figure P13.14, we are to choose which forms noious oides that are removed from automobile ehaust by catalytic converters. We read in Chapter 13 that catalytic converters remove hydrocarbons, CO, and NO gases. Carbon (green) and nitrogen (light blue) form oide gases in automobile engines that are removed by a catalytic converter. In the catalytic converter, hydrocarbons and CO are converted into CO and H O, whereas NO is converted into N and O. 13.15. Collect and Organize Of the highlighted elements in Figure P13.15, we are to identify which are widely used as heterogeneous catalysts. Heterogeneous catalysts have a different phase from the reactants. We read in Section 13.6 about the specific metals used in catalytic converters. Both the transition metals, palladium (blue) and platinum (orange), can serve as heterogeneous catalysts and were identified in the chapter as catalysts. Because catalytic converters contain precious metals such as rhodium, platinum, and palladium, interest in recycling the metals from catalytic converters is great. 13.16. Collect and Organize Of the highlighted elements in Figure P13.16, we are to identify which forms oide radicals (odd-electron species) that catalyze the destruction of the ozone layer. The destruction of ozone in the upper atmosphere is described in Section 13.6.

Chemical Kinetics 45 Free chlorine atoms (yellow) that are generated by UV light interacting with chlorofluorocarbons in the upper atmosphere react with O 3 to form ClO. The average amount of ozone depletion is currently stabilized and the ozone layer is epected to recover by 050, due mostly to the worldwide agreement in the Montreal Protocol to limit the production of chlorofluorocarbons. 13.17. Collect and Organize By considering the levels of O 3 during the day as seen in Figure 13., we are to eplain why [O 3 ] ma occurs later in the day than [NO] ma and [NO ] ma. Ozone in the troposphere (the lowest portion of Earth s atmosphere) is due to the reaction of O with O generated from the interaction of UV light with NO. The presence of NO in the atmosphere and ample sunlight allows the O atoms to react with O to generate O 3. The reactant NO is present in the atmosphere due to automobile ehausts, which build up during the day. The buildup of O 3 lags behind until later in the day, until [NO ] increases and the sunlight becomes stronger as midday approaches. Ozone is a very reactive gas and is irritating to lung tissues. 13.18. Collect and Organize For a plot of the concentration of reactants and products for A B C we are to assess whether [C] ma always appears after [B] ma. From the equation for the reaction, we know that A must react to first form B. Then B can react to form C. Yes, the maimum concentration of C appears after the maimum concentration of B because the formation of C relies on the buildup in concentration of B. Our plot would not necessarily show that all of A has reacted when product C is being formed. 13.19. Collect and Organize We are to eplain why NO concentration does not increase after the evening rush hour. The reaction in the troposphere (lower atmosphere) that produces NO is sunlight NO ( g) NO( g) + O( g ) In the evening the sunlight (and UV radiation) is less intense, so the photochemical breakdown of NO does not occur to as great an etent as after the morning rush hour. The use of catalytic converters to reduce the NO to N and O in automobile ehaust has greatly helped to reduce photochemical smog in large urban and suburban centers.

46 Chapter 13 13.0. Collect and Organize We consider why [O 3 ] reaches a peak in the early afternoon even though O 3 may react with NO to form NO and O. The generation of ozone is due to the photochemical reaction hν NO ( g) NO( g) + O( g) so that O reacts with O to give ozone: ν O( g) + O ( g) h O ( g ) 3 The production of ozone is greatest when the sunlight is intense during the midday. This causes the production of O 3 to be faster than its reaction with NO. The concentration of NO in the atmosphere is epected to be small, so we might epect the rate of its reaction with O 3 to be slow. 13.1. Collect and Organize Using ΔH f for NO, O, and NO in Appendi 4, we can calculate H rn for The ΔH rn can be calculated by using ΔH rn NO(g) + O (g) NO (g) = n products ΔH f,products n products ΔH f,reactants ΔH rn ( ) ( mol NO 90.3 kj/mol) + ( 1 mol O 0.0 kj/mol) = mol NO 33. kj/mol = 114. kj This reaction is eothermic and, therefore, favored by enthalpy. 13.. Collect and Organize Using the enthalpies of formation for O 3, NO, NO, and O gases in Appendi 4, we can calculate the enthalpy of the reaction in which O 3 reacts with NO to produce O and NO. The enthalpy of reaction may be calculated from ΔH rn = n products ΔH f,products n products ΔH f,reactants ΔH rn ( ) + ( 1 mol NO 33. kj/mol) ( ) + ( 1 mol NO 90.3 kj/mol) = 1 mol O 0.0 kj/mol 1 mol O 3 14.7 kj/mol = 199.8 kj This reaction is favored by enthalpy and is eothermic.

Chemical Kinetics 47 13.3. Collect and Organize / For the reaction of N with O to produce N O and N O 5, we are to write balanced chemical equations. 1 (a) N ( g) + O ( g) NO( g) or N ( g) + O ( g) NO( g) 5 (b) N ( g) + O ( g) NO 5( g) or N ( g) + 5 O ( g) N O ( g) 5 Balanced chemical equations usually are written with whole-number coefficients. 13.4. Collect and Organize We are to write balanced chemical equations for the reaction of N O with O to give NO and for the decomposition of N O 5 to NO and O. We can balance these reactions by inspection. 3 (a) NO( g) + O ( g) NO ( g) or NO( g) + 3 O ( g) 4 NO ( g) 1 (b) NO() 5 g NO() g + O( g) or N O ( g) 4 NO ( g) + O ( g) 5 NO itself is reactive because it is a radical species with an odd number of electrons. 13.5. Collect and Organize We are to eplain the difference between the average rate and the instantaneous rate of a reaction. The rate of reaction is measured from the change in concentration of a reactant or product over time. The difference between the average and instantaneous rates is the length of time over which the change in concentration is measured. The average rate is the rate averaged over a fairly long time, whereas the instantaneous rate is the rate at a specific moment in time (or over a very, very short time).

48 Chapter 13 The rate of a reaction is always positive. 13.6. Collect and Organize We consider whether the average and the instantaneous rate may be equal. The average rate is the rate averaged over a fairly long period, whereas the instantaneous rate is the rate at a specific moment in time (or over a very, very short period). Yes, the average rate and the instantaneous rate may be the same, as long as the points in time over which the average rate is measured are chosen so that the slope of the average rate line is equal to the slope of the tangent to the line for the instantaneous rate. Also, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. 13.7. Collect and Organize We are to eplain why the average rates of most reactions change over time. The forward rate of a reaction as measured from the average rate depends on the concentration of reactants. As the reaction proceeds, the concentrations of the reactants decrease. Because most reactions depend on the availability (that is, concentration) of reactants to proceed, the decrease in reactant concentrations lowers the reaction rate.

Chemical Kinetics 49 Reactions that have no dependence on the concentrations of the reactants, although rare, do not show a change in rate as the reaction proceeds. 13.8. Collect and Organize We consider whether the instantaneous rate of a reaction changes with time. The instantaneous rate is the rate of the reaction measured at a specific moment in time. It is the absolute value of the slope of the line that is tangent to the curve of reactant or product concentration versus time. Yes, for most reactions, the instantaneous rate changes (decreases) with time until the reaction reaches equilibrium. When we study chemical equilibria in Chapter 16, we will learn that at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. 13.9. Collect and Organize Given the balanced equation for the reaction of ammonia with oygen to produce H +, NO, and H O, we are asked to relate the formation of products and consumption of reactants. The coefficients in the balanced chemical equation tell us how the rate of formation of products and consumption of reactants are related to each other. For the generic chemical equation where A, B, C, and D are the reactants and products and a, b, c, and d are their stoichiometric coefficients, a A + b B c C + d D the relationship of the rates is given by For the reaction given Rate = 1 Rate = 1 a Δ NH 3 = 1 3 Δ A = 1 b Δ O = 1 Δ B = 1 c Δ H + = 1 Δ C = 1 d Δ NO Δ D = 1 Δ H O

50 Chapter 13 (a) [ ] + Δ NH 3 Δ H Δ NO = =, the rate of consumption of NH 3, is equal to the rate of formation of H + and NO. Δ NO [ O ] (b) Δ =, the rate of formation of NO, is two-thirds the rate of consumption of O. 3 Δ[ NH3] Δ[ O] (c) =, the rate of consumption of NH 3, is two-thirds the rate of consumption of O. 3 The negative sign is used in front of the epressions involving the consumption of a reactant to give a positive reaction rate because the change in concentration, [X] f [X] i, or Δ[X], is negative for reactants because [X] f < [X] i. 13.30. Collect and Organize Using the balanced equation describing the reaction of CO with NO, we are to write epressions to compare the rates of formation of products and the rates of consumption of reactants. From the balanced equation, we see that the reaction may be epressed as Rate = Δ N = 1 Δ CO = 1 Δ CO = 1 Δ NO Δ[ N ] 1 Δ[ CO] (a) Because =, the rate of formation of N is one-half the rate of consumption of CO. Δ t Δ t 1 Δ[ CO ] 1 Δ[ NO] (b) Because =, the rate of formation of CO is the equal to the rate of consumption of NO. 1 Δ[ CO] 1 Δ[ NO] (c) Because =, the rate of consumption of the two reactants is equal. These relative rates make sense according to the stoichiometry of the reaction. For every mol of CO and NO used in the reaction, mol of CO and 1 mol of N are produced. So, for eample, the concentration of CO decreases at twice the rate that the concentration of N increases. 13.31. Collect and Organize For each of three reactions, we are to write epressions for the rate of formation of products and consumption of reactants. The coefficients in the balanced chemical equation tell us how the rate of formation of products and consumption of reactants are related to each other. For the generic chemical equation where A, B, C, and D are the reactants and products and a, b, c, and d are their stoichiometric coefficients, a A + b B c C + d D the relationship of the rates is given by Rate = 1 a Δ A = 1 b Δ B = 1 c Δ C = 1 d Δ D

Chemical Kinetics 51 Δ[ HO ] 1 Δ[ OH] (a) Rate = = Δ[ ClO] Δ[ O] Δ[ ClO3] (b) Rate = = = Δ N O Δ H O 1 Δ HNO (c) Rate = = = [ 5] [ ] [ 3] The negative sign is used in front of the epressions involving the consumption of a reactant to give a positive reaction rate because the change in concentration, [X] f [X] i, or Δ[X], is negative for reactants because [X] f < [X] i. 13.3. Collect and Organize For each reaction we are to write an epression relating the rate of formation of products to the rate of consumption of reactants. In the epression relating the rate of a reaction in terms of the formation of products and consumption of reactants, the denominator of the fraction for each product s or reactant s change of concentration versus time is the coefficient of that product or reactant in the overall balanced equation: a A + b B = c C + d D (a) Rate = Δ Cl O = 1 Δ ClO (b) Rate = Δ N O 5 (c) Rate = 1 Δ INO = Δ NO = Δ I = 1 Rate = 1 a = Δ NO 3 Δ NO ΔA = 1 b ΔB = 1 c ΔC = 1 d ΔD The negative sign is used in front of the epressions involving the consumption of a reactant to give a positive reaction rate because the change in concentration, [X] f [X] i, or Δ[X], is negative for reactants because [X] f < [X] i. 13.33. Collect and Organize Using the balanced equation describing the reaction of SO with CO, we are to write epressions to compare the rates of formation of products and the rates of consumption of reactants. From the balanced equation, we see that the reaction may be epressed as Δ CO Δ CO (a) Rate = = 3 [ ] [ ] Rate = Δ SO = 1 3 Δ CO = 1 Δ CO = Δ COS

5 Chapter 13 [ ] [ ] Δ COS Δ SO (b) Rate = = Δ[ CO] Δ[ SO ] (c) Rate = = 3 These relative rates make sense according to the stoichiometry of the reaction. For every 1 mol of SO used in the reaction, mol of CO and 1 mol of COS are produced. So, for eample, the concentration of CO will increase twice as fast as the concentration of SO decreases. 13.34. Collect and Organize Using the balanced equation describing the reaction of CH 4 with NO, we are to write epressions to compare the rates of formation of products and the rates of consumption of reactants. From the balanced equation, we see that the reaction may be epressed as Rate = Δ CH 4 = 1 4 Δ[ N] Δ[ CO] (a) Rate = = Δ[ CO ] 1 Δ[ NO] (b) Rate = = 4 Δ CH4 1 Δ HO (c) Rate = = [ ] [ ] Δ NO = 1 Δ N = Δ CO = 1 Δ H O The negative sign is used in front of the epressions involving the consumption of a reactant to give a positive reaction rate because the change in concentration, [X] f [X] i, or Δ[X], is negative for reactants because [X] f < [X] i. 13.35. Collect and Organize Using the relative rate epressions and the rate of the consumption of ClO in two reactions, we are to calculate the rate of change in the formation of the products of the two reactions. (a) For this reaction, (b) For this reaction, Rate = 1 Rate = Δ ClO Δ ClO = Δ O 3 = Δ Cl = Δ O = Δ O = Δ ClO For each reaction we are given [ClO]/ t. We can use this value in the relationships above to calculate the rate of change in (a) the concentration of Cl and O and (b) the concentration of O and ClO. (a) Rate = Δ Cl = Δ O = 1 = 1. 10 7 M/s Δ ClO = 1.3 107 M/s

Chemical Kinetics 53 (b) Rate = Δ O =.9 10 4 M/s = Δ ClO = Δ ClO = (.9 10 4 M/s) The rates of the formation of products are positive because [X] f > [X] i, so [X] f [X] i = Δ[X] is positive. 13.36. Collect and Organize Using the relative rate epressions and the given rates of the consumption of NO 3 in two reactions, we can calculate the rates of change in the formation of the products of the two reactions. (a) For this reaction, Rate = Δ NO 3 Δ NO 3 = Δ NO = 1 Δ NO Δ NO = Δ O (b) For this reaction, Rate = 1 = 1 For each reaction we are given [NO 3 ]/ t. We can use this value in the rate relationships to calculate the rates of change in the concentration of NO. (a) 1 Δ NO = Δ NO 3 or ( ) =. 10 5 mm/min = 4.4 10 5 mm/min Δ NO = Δ NO 3 (b) 1 Δ NO = 1 Δ NO 3 ( ) =.3 mm/min =.3 mm/min or Δ NO = Δ NO 3 Because we always epress the rate of reaction as a positive value, we assign the rate of consumption of a reactant a negative sign because [X] f [X] i < 0, since [X] f < [X] i. 13.37. Collect and Organize Given the [O 3 ] over time when it reacts with NO, we are to calculate the average reaction rate for two intervals. The average rate of reaction can be found according to Δ O 3 = O 3 f O 3 t f t i i Between 0 and 100 µs: Δ O 3 ( ) M ( ) µs = 9.93 10 3 1.13 10 100 0 = 1.4 10 5 M/µs

54 Chapter 13 Between 00 and 300 µs: Δ O 3 ( ) M ( ) µs = 8.15 10 3 8.70 10 3 300 00 = 5.50 10 6 M/µs As the reaction proceeds, the rate of consumption of ozone decreases. This is due to the decreasing reactant concentrations. 13.38. Collect and Organize Given the [N O 5 ] over time when it decomposes, we are to calculate the average rate of the reaction between all the consecutive times for the data presented. The average rate of the reaction can be found according to Δ NO NO NO = t t [ ] [ 5] [ 5] 5 f i Between 0 and 1.45 s: 1 1 3 Δ[ NO] ( 1.357 10 1.500 10 5 ) molecules/cm 10 3 = = 9.86 10 molecules (cm s) ( 1.45 0 ) s Between 1.45 s and.90 s: 1 1 3 Δ[ NO] ( 1.8 10 1.357 10 5 ) molecules/cm 10 3 = = 8.90 10 molecules (cm s) (.90 1.45) s Between.90 and 4.35 s: 1 1 3 Δ[ NO] ( 1.111 10 1.8 10 5 ) molecules/cm 10 3 = = 8.07 10 molecules (cm s) ( 4.35.90 ) s Between 4.35 and 5.80 s: 1 1 3 Δ[ NO 5] ( 1.005 10 1.111 10 ) molecules/cm 10 3 = = 7.31 10 molecules (cm s) 5.80 4.35 s ( ) As the reaction proceeds, the rate of disappearance of dinitrogen pentoide decreases. This is due, in part, to the decreasing reactant concentrations. 13.39. Collect and Organize After we plot [ClO] versus time and [Cl O ] versus time, we can determine the instantaneous rate of change at 1 s for each compound. The instantaneous rate is the slope of the line that is tangent to the curve at the time we are interested in. We can estimate fairly well the instantaneous rate at 1 s from the plots by choosing two points that are close to 1 s to calculate the slope. Since we are given only values for [ClO], we need to calculate [Cl O ] for each time. ( [Cl O ] t = ClO 0 ClO t ) where [ClO] 0 =.60 10 11 M f i

Chemical Kinetics 55 Time, s [ClO] t, molecules/cm 3 [Cl O ] t, molecules/cm 3 0.60 10 11 0.00 1 1.08 10 11 7.60 10 10 6.83 10 10 9.59 10 10 3 4.99 10 10 1.05 10 11 4 3.93 10 10 1.10 10 11 5 3.4 10 10 1.14 10 11 6.76 10 10 1.16 10 11 Initially, no Cl O is present, so [Cl O ] 0 = 0 molecules/cm 3. For the change in concentration of ClO versus time, we obtain the following plot: To get a fairly good estimate from the data given for the instantaneous rate, we can choose two points from the data set that surrounds the data point of interest and calculate [ClO]/ t. For t = 1 s, we can use the points t = 0 s and t = s: 10 11 3 Δ [ClO] (6.83 10.60 10 ) molecules/cm 10 3 1 = = 9.59 10 molecules cm s ( 0) s Using a graphing program that calculates the slope of the tangent to the line at t = 1 s, we get an instantaneous rate of 8.8 10 10 molecules cm 3 s. For the change in concentration of Cl O versus time, we obtain the following plot.

56 Chapter 13 To get a fairly good estimate from the data given for the instantaneous rate, we can choose two points from the data set that surrounds the data point of interest and calculate [Cl O ]/ t. For t = 1 s, we can use the points t = 0 s and t = s: 10 3 Δ [ClO ] (9.59 10 0) molecules/cm 4.80 10 10 molecules cm 3 s 1 = = ( 0) s Using a graphing program that calculates the slope of the tangent to the line at t = 1 s, we get an instantaneous rate of 4.13 10 10 molecules cm 3 s. Because we epect the rate of disappearance of ClO to be twice the rate of appearance of Cl O from the balanced equation ClO(g) Cl O (g) our answers make sense: Δ[ClO] 10 3 1 9.59 10 molecules cm s = =.0 10 3 1 Δ [ClO ] 4.80 10 molecules cm s 13.40. Collect and Organize To determine the instantaneous rate of the reaction of O 3 with NO at 0.000 s and 0.05 s, we have to plot [NO] versus time. On the plot, we will draw the tangent to the line where t = 0.00 s and t = 0.05 s. The slopes of these lines are the instantaneous rates of the reaction. We can estimate fairly well the instantaneous rate at the two times during the reaction from the plot by choosing two points from the data set given that surrounds those times of interest to calculate the slope. To get a fairly good estimate from the data given for the instantaneous rate, we can choose two points from the data set that surrounds the data point of interest and calculate [NO]/ t. For t = 0.00, we can use the points t = 0.00 and t = 0.011: 8 8 Δ [NO] (1.8 10.0 10 ) M 7 = = 1.8 10 M/s (0.011 0.000) s Using a graphing program that calculates the slope of the tangent to the line at t = 0.00 s, we get an instantaneous rate of 1.9 10 7 M/s.

Chemical Kinetics 57 For t = 0.05, we can use the points where t = 0.07 s and t = 0.10 s: 8 8 Δ [NO] (1. 10 1.6 10 ) M 8 = = 5.33 10 M/s (0.10 0.07) s Using a graphing program that calculates the slope of the tangent to the line at t = 1 s, we get an instantaneous rate of 6.7 10 8 M/s. As the reaction progresses, the rate of the reaction decreases. 13.41. Collect and Organize We consider whether two different chemical reactions can have the same rate law epression. The rate law epression is of the form Rate = k[a] where k is the rate constant, [A] is the concentration of the reactant(s) on which the rate depends, and is the order of the reaction for that concentration of reactant. Yes, two different reactions can have the same form of the rate law. They both may have the same dependence on the concentration of reactants, yet yield different products. For eample, A may decompose by two routes: A B + C A D + E If both reactions are first order (or even second order) in [A], they would have the same rate law: Rate = k[a] The value of the rate constant, however, is epected to be different for the two reactions. 13.4. Collect and Organize We are to eplain why the units of the rate constant change with the overall order of the reaction. The form of the rate law is Rearranging and solving for k: Rate = k[a] k = Rate A The rate of a reaction is epressed as concentration per time (often in moles per liter per second) and the concentration of A is usually epressed in moles per liter (M ). The order of the reaction may be any number (fractional, whole numbers, or even negative). The order of the reaction changes the units of k. Some eamples are as follows: M s 1 For a first-order reaction, k = = s M M s 1 For a third-order reaction, k = = M s 3 M For a one-half-order reaction, k = M s M = M 1/ s 1 1/ Once the form of the rate law is known, we can determine the units of the rate constant.

58 Chapter 13 13.43. Collect and Organize We are asked whether the units of the half-life for a second-order reaction are the same as those of the half-life for a first-order reaction. The half-life is the time for the amount of reactant originally present to decrease by one-half. Yes. Because the half-life is a time measurement, the units, no matter what the order of the reaction, are always in units of time (s, min, hr, yr, etc.). The half-life of a reaction depends on the value of the reaction s rate constant and, ecept for first-order reactions, on the initial concentration. The larger the rate constant, the faster the reaction and the shorter the half-life of the reaction. 13.44. Collect and Organize We consider whether the half-life of a first-order reaction depends on the concentration of the reactants. For a first-order reaction, t 1/ 0.693 = k No, there is no dependence of t 1/ for a first-order reaction on the concentration of reactants because [A] 0 does not appear in the epression for the half-life. The half-life for a first-order reaction is inversely related to the rate constant, k. 13.45. Collect and Organize For a second-order reaction, we are to predict the effect of doubling [A] 0 on the half-life. For a second-order reaction, t 1/ = k 1 [ A] 0 From the equation for the half-life of a second-order reaction, we see that doubling [A] 0 halves the half-life. The half-life of a second-order reaction, like that of a first-order reaction, is inversely related to the rate constant. 13.46. Collect and Organize For two decomposition reactions that have the same rate constant, k, we are to determine whether these reactions will also have the same t 1/. The half-life of a first-order reaction is t 1/ 0.693 = k

Chemical Kinetics 59 Because t 1/ for a first-order reaction is dependent only on the value of k, yes, two reactions with the same rate constant have the same half-life. The reactions may not necessarily have the same value of k at a different temperature, though. 13.47. Collect and Organize For each rate law epression, we are to determine the order of the reaction with respect to each reactant and the overall reaction order. The order of a reaction is the eperimentally determined dependence of the rate of a reaction on the concentration of the reactants involved in the reaction. In the rate law epression, the order is shown as the power to which the concentration of a particular reactant is raised. The overall reaction order is the sum of the powers of the reactants in the rate law epression. (a) For the rate law epression Rate = k[a][b], the reaction is first order in both A and B and second order overall. (b) For Rate = k[a] [B], the reaction is second order in A, first order in B, and third order overall. (c) For Rate = k[a][b] 3, the reaction is first order in A, third order in B, and fourth order overall. The higher the order of the reaction for a particular reactant, the greater the effect of a change in concentration of that reactant on the reaction rate. 13.48. Collect and Organize For each rate law epression, we are to determine the order of the reaction with respect to each reactant and the overall reaction order. The order of a reaction is the eperimentally determined dependence of the rate of a reaction on the concentration of the reactants involved in the reaction. In the rate law epression, the order is shown as the power to which the concentration of a particular reactant is raised. The overall reaction order is the sum of the powers of the reactants in the rate law epression. (a) For the rate law epression Rate = k[a] [B] 1/, the reaction is second order in A, one-half order in B, and two-an-one-half order overall. (b) For Rate = k[a] [B][C], the reaction is second order in A, first order in both B and C, and fourth order overall. (c) For Rate = k[a][b] 3 [C] 1/, the reaction is first order in A, third order in B, one-half order in C, and four-andone-half order overall. The higher the order of the reaction for a particular reactant, the greater the effect of a change in concentration of that reactant on the reaction rate. 13.49. Collect and Organize For each reaction described, we are to write the rate law and determine the units for k, using the units M for concentration and s for time. The general form of the rate law is Rate = k[a] [B] y

60 Chapter 13 where k is the rate constant, A and B are the reactants, and and y are the orders of the reaction with respect to each reactant as determined by eperiment. (a) Rate = k[o][no ] Because rate has units of M/s and each concentration has units of M, M s 1 1 k = = M s M (b) Rate = k[no] [Cl ] Because rate has units of M/s and each concentration has units of M, M s 1 k = = M s 3 M (c) Rate = k [CHCl 3 ][Cl ] 1 Because rate has units of M/s and each concentration has units of M, M s 1 1 k = = M s 3/ M (d) Rate = k[o 3 ] [O] 1 Because rate has units of M/s and each concentration has units of M, M s 1 k = = s M M The units of the rate constant clearly depend on the overall order of the reaction. 13.50. Collect and Organize Given the order of the reaction for each of the reactants in the chemical reaction A + B C we are to write the rate law epression and determine the units of the rate constant k, using the units M for concentration and s for time. In the rate law, the order of the reaction for each reactant is placed as an eponent to the concentration: Rate = k[a] [B] y where and y are the orders of the reaction with respect to A and B. The rate of the reaction is epressed as M/s, so the rate constant units will have to be 1/M z 1 s, where z is the overall order of the reaction. (a) Rate = k[a][b], k in units of 1/M s (b) Rate = k[a][b], k in units of 1/M s (c) Rate = k[b], k in units of 1/M s (d) Rate = k[a] [B], k in units of 1/M 3 s A reaction rate that is independent of the concentration of a reactant is zero order in that reactant. So for part c, Rate = k[a] 0 [B] where [A] 0 = 1. 13.51. Collect and Organize Given the changes in rate of the decomposition of BrO to Br and O when [BrO] is changed, we are to predict the rate law in each case.

Chemical Kinetics 61 The general form of the rate law for the reaction is Rate = k[bro] where is an eperimentally determined eponent. (a) If the rate doubles when [BrO] doubles, then = 1 and the rate law is Rate = k[bro] (b) If the rate quadruples when [BrO] doubles, then = and the rate law is Rate = k[bro] (c) If the rate is halved when [BrO] is halved, then = 1 and the rate law is Rate = k[bro] (c) If the rate is unchanged when [BrO] is doubled, then = 0 and the rate law is Rate = k[bro] 0 = k For this reaction, the relationship is straightforward between the change in rate when the concentration of the reactant was changed to determine. To determine for more complicated reactions, use rate A = rate 1 A 1 ln rate rate 1 = ln A A ln rate rate = 1 A ln A 1 13.5. Collect and Organize Using the information of how the rate of the reaction changes when the concentration of a reactant is changed, we can predict the rate law for the reaction of NO with Br to give NOBr. (a) If the rate doubles when [NO] doubles while [Br ] remains constant, then the reaction is first order in NO. (b) If the rate doubles when [Br ] doubles while [NO] remains constant, then the reaction is first order in Br. (c) If the rate of the reaction increases by 1.56 times when [NO] increases by 1.5 times while [Br ] remains constant, then the order of the reaction for NO is 1.5 = 1.56 ln(1.5) = ln(1.56) = 1.993 = (d) If the rate is halved when [NO] is doubled and [Br ] remains constant, then the order of the reaction of NO is = 0.5 ln() = ln(0.5) = 1 (a) Rate = k[no][br ] y, where y is undefined (b) Rate = k[no] y [Br ], where y is undefined 1

6 Chapter 13 (c) Rate = k[no] [Br ] y, where y is undefined (d) Rate = k[br ] y /[NO], where y is undefined The order for a particular reactant in a rate law epression may be any number (positive, negative, zero, fraction, or whole) and can be determined only by eperiment, not by looking at the overall balanced chemical equation. 13.53. Collect and Organize Given that the rate of the reaction quadruples when both [NO] and [ClO] are doubled, we are to identify what additional information we would need to write the rate law for the reaction NO(g) + ClO(g) NO (g) + Cl(g) The general form of the rate law for this reaction is Rate = k[no] [ClO] y If the rate quadruples when both [NO] and [ClO] are doubled, the rate law could be any of the following: Rate = k[no][clo] Rate = k[no] Rate = k[clo] To differentiate among these, we need to determine the change in the rate when only [NO] or [ClO] is changed. If the rate is only doubled when [NO] and [ClO] are independently changed, then the rate law is Rate = k[no][clo] If the rate is quadrupled when [NO] is doubled but remains constant if [ClO] is doubled, then the rate law is Rate = k[no] If the rate is quadrupled when [ClO] is doubled but remains constant if [NO] is doubled, then the rate law is Rate = k[clo] 13.54. Collect and Organize Given the information that the rate for the reaction between M (a molecule that is unchanged in the reaction) with ClO and NO is first order in NO and in ClO, we are to write the rate law and determine the order of the reaction for M. Knowing that the reaction is first order in both NO and ClO means that the eponent for the concentrations in the rate law for both of these reactants is 1. Because no dependence on the rate was given in the problem for M, we can assume that changing [M] does not affect the rate of the reaction. (a) Rate = k[clo][no] (b) For the rate law to show no dependence on [M], the order of the reaction with respect to M must be zero. The molecule M may function in this reaction to transfer energy from one of the reactants to another to enable the reaction to take place or to bring the reactants closer together to facilitate the reaction between them. 13.55. Collect and Organize For the reaction of NO with O 3 to produce NO 3 and O, we are to write the rate law given that the reaction is first order in both NO and O 3. From the rate law and given the rate constant, we can calculate the rate of the

Chemical Kinetics 63 reaction for a given [NO ] and [O 3 ]. From this we can calculate the rate of appearance of NO 3 and the rate of the reaction when [O 3 ] is doubled. The general form of the rate law for this reaction is Rate = k[no ] [O 3 ] y The rate of consumption of reactants and formation of products is Rate = Δ NO = Δ O 3 (a) Rate = k[no ][O 3 ] 4 1.93 10 8 7 11 (b) Rate = 1.8 10 M 1.4 10 M = 4.9 10 M/s M s Δ[ NO ] (c) Rate = 3 = 4.9 10 11 M/s (d) When [O 3 ] is doubled, the rate of the reaction doubles. = Δ NO 3 = Δ O When [O 3 ] =.8 10 7 M (double that in part b) the rate of reaction is 9.73 10 11 M/s, which is twice that calculated in part b, so our prediction in part d is correct. 13.56. Collect and Organize We are to write the rate law for the reaction between N O 5 and H O, and from that rate law we are to calculate the rate constant for the reaction for a specified reaction rate and concentrations of N O 5 and H O. (a) We are given that the reaction is first order in both reactants, so the eponents for the concentrations of both N O 5 and H O are 1. (b) To calculate k we rearrange the rate law to solve for k and use the given rate of the reaction and the concentrations of the reactants in the equation. (a) Rate = k[n O 5 ][H O] 4 rate 4.55 10 m M/min (b) k = = = 1.50 10 mm min [N O ][H O] 0.13 mm 30 mm 5 5 1 1 The rate constant does not change (ecept with temperature), although changing the initial concentrations of the reactants would change the rate of the reaction. 13.57. Collect and Organize By comparing the rate constants for four reactions that are all second order, we can determine which reaction is the fastest if all the initial concentrations are the same. The reaction with the largest rate constant has the fastest reaction rate. Reaction (c) has the largest value of k, so it proceeds the fastest. The slowest reaction is a reaction with the smallest value of k.

64 Chapter 13 13.58. Collect and Organize We are asked to compare a first-order reaction with a second-order reaction, each with the same magnitude of rate constant, k. The first-order reaction rate law is The second-order reaction rate law is Rate A = k[a] Rate B = k[b] (a) Neither. If the initial concentrations of A and B are both 1.0 mm, the reaction rates are identical because Rate A = k[1.0 mm] Rate B = k[1.0 mm] Rate A = Rate B (b) If the initial concentrations of A and B are both.0 M, the reaction rates differ: Rate A = k[.0 M] Rate B = k[.0 M] = k 4.0 M For these concentrations of reactants, the second-order reaction proceeds faster. For these two reactions, any [A] = [B] less than 1.0 mm will have a faster rate for the first-order reaction, but for [A] = [B] greater than 1.0 mm the second order-reaction will be faster. 13.59. Collect and Organize Given the information that the rate of the reaction between NO and NO with water doubles when either [NO] or [NO ] doubles and the rate does not depend on [H O], we can write the rate law for the reaction. The general form of the rate law for this reaction is Rate = k[no] [NO ] y [H O] z Doubling of the reaction rate with doubling of [NO] or [NO ] means the reaction is first order in those reactants. Because the rate does not depend on [H O], the reaction is zero order in that reactant. The rate law for this reaction is Rate = k[no] 1 [NO ] 1 [H O] 0 = k[no][no ] If [NO] and [NO ] are doubled simultaneously, the rate of reaction quadruples. 13.60. Collect and Organize Given the effect on the rate of the reaction between HO and SO 3 when their concentrations are doubled, we can write the rate law for the reaction. If the rate of the reaction doubles when either reactant concentration is doubled, then the eponent for each reactant in the rate law epression is 1. Rate = k[ho ][SO 3 ] Although this reaction is first order in each reactant, it is second order overall.

Chemical Kinetics 65 13.61. Collect and Organize In the reaction of ClO with OH the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two eperiments in which the concentration of that reactant changes but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law epression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the eperiments. Using eperiments 1 and, we find that the order of the reaction with respect to ClO is 1: rate [ClO 1 ] 0,1 = rate [ClO ] 0, 0.048 M/s 0.060 M = 0.0087 M/s 0.00 M 3.00 = 3.01 = 1 Using eperiments and 3, we find that the order of the reaction with respect to [OH ] is also 1: rate [OH ] 3 0,3 = rate [OH ] 0, 0.047 M/s 0.090 M = 0.0087 M/s 0.030 M.99 = 3.0 = 1 The rate law for this reaction is Rate = k[clo ][OH ]. Rearranging the rate law epression to solve for k and using the data from eperiment 1 gives rate 0.048 M/s 1 1 k = 14 s [ClO ][OH ] = M 0.060 M 0.030 M = We may use any of the eperiments in the table to calculate k. Each eperiment s data give the same value of k as long as the eperiments were all run at the same temperature. 13.6. Collect and Organize In the reaction of NO with O 3, the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two eperiments in which the concentration of that reactant changes but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law epression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the eperiments.

66 Chapter 13 Using eperiments 1 and, we find that the order of the reaction with respect to NO is 1: rate [NO ] 0, = rate 1 [NO ] 0,1 37.5 M/s 0.0150 M = 5 M/s 0.0100 M 1.5 = 1.50 = 1 Using eperiments 3 and 4, we find that the order of the reaction with respect to O 3 is also 1: rate [O 4 3] 0,4 = rate 3 [O 3] 0,3 00.0 M/s 0.000 M = 50.0 M/s 0.0050 M 4.00 = 4.0 = 1 The rate law for this reaction is Rate = k[no ][O 3 ]. Rearranging the rate law epression to solve for k and using the data from eperiment 1 gives rate 5 M/s 5 1 1 k = 5.0 10 s [NO ][O ] = M 0.0100 M 0.0050 M = 3 We may use any of the eperiments in the table to calculate k. Each eperiment s data give the same value of k as long as the eperiments were all run at the same temperature. 13.63. Collect and Organize In the reaction of H with NO, the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two eperiments in which the concentration of that reactant changes, but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law epression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the eperiments. Using eperiments 1 and, we find that the order of the reaction with respect to NO is : rate [NO] 0, = rate 1 [NO ] 0,1 0.0991 M/s 0.7 M = 0.048 M/s 0.136 M 4.00 =.00 =

Chemical Kinetics 67 Using eperiments 3 and 4, we find that the order of the reaction with respect to H is 1: rate [H 4 ] 0,4 = rate 3 [H ] 0,3 1.59 M/s 0.848 M = 0.793 M/s 0.44 M.01 =.00 = 1 The rate law for this reaction is Rate = k[no] [H ]. Rearranging the rate law epression to solve for k and using the data from eperiment 1 gives 0.048 M/s 1 k = = 6.3 M s 0.136 M 0.1 M ( ) We may use any of the eperiments in the table to calculate k. Each eperiment s data give the same value of k as long as the eperiments were all run at the same temperature. 13.64. Collect and Organize In the reaction of NO with CO, the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. We are also to calculate the rate of appearance of CO when [NO ] = [CO] = 0.500 M. To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two eperiments in which the concentration of that reactant changes but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law epression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the eperiments. Finally we can calculate the rate of the formation of a product from the concentrations of the reactants by substituting those concentrations and the calculated value of k into the rate law epression, making sure the result agrees with the stoichiometry of the reaction. (a) Using eperiments 1 and, we find that the order of the reaction with respect to CO is 0: rate [CO] 0, = rate 1 [CO] 0,1 5 1.44 10 M/s 0.413 M = 5 1.44 10 M/s 0.86 M 1.00 = 0.500 = 0 Using eperiments and 3, we find that the order of the reaction with respect to NO is : rate [NO 3 ] 0,3 = rate [NO ] 0, 5 5.76 10 M/s 0.56 M = 5 1.44 10 M/s 0.63 M 4.00 =.00 = The rate law for this reaction is Rate = k[co] 0 [NO ] = k[no ].

68 Chapter 13 (b) Rearranging the rate law epression to solve for k and using the data from eperiment 1 gives 5 rate 1.44 10 M/s 4 1 1 k = = =.08 10 M s [NO ] (0.63 M ) (c) The rate of formation of CO when [NO ] = [CO] = 0.500 M is.08 10 4 Rate = (0.500 M ) = 5.0 10 5 M /s M s Because the concentration of CO has no effect on the rate of the reaction, [CO] has an eponent of zero and the carbon monoide concentration does not appear in the rate law epression. 13.65. Collect and Organize From the data given for the concentration of NO 3 over time as it decomposes to NO and O, we are to calculate the value of k for this reaction. We are given a single data set and the fact that the reaction is second order. For this second-order reaction the plot of 1/[NO 3 ] versus time gives a straight line with a slope equal to k, the reaction rate constant. 9.00E+0 1/[NO 3 ] (µm 1 ) 8.50E+0 8.00E+0 7.50E+0 7.00E+0 y = 0.30 + 680.4 R² = 1 6.50E+0 6.00E+0 0 100 00 300 400 500 600 Time (min) The slope = k = 0.3 µm 1 min 1. The half-life of this second-order reaction is 1 1 t 1/ = = k[no 3 ] 0 0.30 µm 1 min 1 = 14 min, or 35.4 h ( ) ( 1.470 10 3 µm ) 13.66. Collect and Organize From the data given for the concentration of ClOO over time as it decomposes to Cl and O, we are to determine the rate law and calculate the value of k for this reaction. We are given a single data set. If a plot of [ClOO] versus time yields a straight line, the reaction is zero order. If the plot of ln[cloo] versus time yields a straight line, the reaction is first order. If a plot of 1/[ClOO] versus time yields a straight line, the reaction is second order. The slope of the line on the appropriate graph gives the rate constant k for the reaction (slope = k for a zero- or first-order reaction and slope = +k for a second-order reaction).

Chemical Kinetics 69 [ClOO] (M).00E-06 1.50E-06 1.00E-06 5.00E-07 0.00E+00-5.00E-07 y = 5E 07 + 1E 06 R² = 0.587 0 0.5 1 1.5.5 3 Time (µs) Zero-Order Plot ln[cloo] (M) -10-1 -14-16 -18-0 - -4-6 -8-30 y = 3.015 13.15 R² = 0.99987 0 0.5 1 1.5.5 3 Time (µs) First-Order Plot 1/[ClOO] (1/M) 3.00E+09.50E+09.00E+09 1.50E+09 1.00E+09 5.00E+08 0.00E+00-5.00E+08-1.00E+09 y = 8E+08 5E+08 R² = 0.6101 0 0.5 1 1.5.5 3 Time (µs) Second-Order Plot The first-order plot is linear with the slope = k = 3.0, so k = 3.0 µs 1. The rate law is Rate = k[cloo]. The half-life of this first-order reaction is t 1/ = 0.693 k = 0.693 = 0.3 µs 1 3.0 µs 13.67. Collect and Organize From the data given for concentration of NH 3 over time as it decomposes to N and H, we are to determine the rate law and calculate the value for k for this reaction. We are given a single data set. If a plot of [NH 3 ] versus time yields a straight line, the reaction is zero order. If a plot of ln[nh 3 ] versus time yields a straight line, the reaction is first order. If a plot of 1/[NH 3 ] versus time yields a straight line, the reaction is second order. The slope of the line on the appropriate graph gives the rate constant k for the reaction (slope = k for a zero- or first-order reaction and slope = +k for a second-order reaction).

70 Chapter 13 [NH 3 ] (M) 3.00E-0.50E-0.00E-0 1.50E-0 1.00E-0 y = 3.1E 05 +.36E 0 R² = 9.45E 01 Zero-Order Plot ln[nh 3 ] -3-3.5-4 -4.5-5 First-Order Plot y = 3.00E 03 3.67E+00 R² = 1.00E+00 5.00E-03-5.5 0.00E+00 0 00 400 600 800 Time (s) -6-6.5 0 00 400 600 800 Time (s) Second-Order Plot 1/[NH 3 ] 400.00 350.00 300.00 50.00 00.00 150.00 100.00 50.00 0.00 y = 0.408 + 0.16 R² = 0.935 0 00 400 600 800 Time (s) The first-order plot is linear, so that the rate law is Rate = k[nh 3 ]. The slope = k = 0.003 s 1 and thus k = 0.003 s 1. The half-life of this first-order reaction is t 1 = 0.693 0.693 = = 31 s, or 3.9 min 1 k 0.0030 s 13.68. Collect and Organize From the data given for the concentration of HO over time as it decomposes to H O and O, we are to determine the rate law and calculate the value of k for this reaction. We are given a single data set. If a plot of [HO ] versus time yields a straight line, the reaction is zero order. If a plot of ln[ho ] versus time yields a straight line, the reaction is first order. If a plot of 1/[HO ] versus time yields a straight line, the reaction is second order. The slope of the line on the appropriate graph gives the rate constant k for the reaction (slope = k for a zero- or first-order reaction and slope = +k for a second-order reaction).

Chemical Kinetics 71 10 Zero-Order Plot.5 First-Order Plot [HO ] (µm) 8 6 4 y = 3.0054 + 7.3899 R² = 0.90948 ln[ho ] 1.5 1 y = 0.8543 +.1399 R² = 0.99985 0.5 0 0 0.5 1 1.5.5 3 Time (µs) 0 0 0.5 1 1.5.5 3 Time (µs) 1/[HO ] 1.00 0.80 0.60 0.40 y = 0.35 + 0.0198 R² = 0.9119 Second-Order Plot 0.0 0.00 0 0.5 1 1.5.5 3 Time (µs) The first-order plot is linear with the slope = k = 0.86, so k = 0.86 µs 1. The rate law is Rate = k[ho ]. The half-life of this first-order reaction is t 1/ = 0.693 k = 0.693 = 0.81 µs 1 0.86 µs 13.69. Collect and Organize For the decomposition reaction of N O 5 to NO and O, we are given that the reaction is first order with a rate constant, k, equal to 6.3 10 4 s 1. We are to calculate the amount of N O 5 that remains after 1 hr of reaction time when the initial concentration of N O 5 is 0.50 mol/l and determine the percentage of N O 5 that has reacted. The integrated rate law for a first-order reaction is ln[a] = kt + ln[a] 0 The concentration of N O 5 remaining after 1 hr is ln[n O 5 ] = kt + ln[n O 5 ] 0 6.3 10 4 60 min ln[n O 5 ] = s 1 h 1 h 60 s 1 min + ln 0.50 mol/l [N O 5 ] = e.9683 = 0.051 mol/l The percentage of N O 5 reacted is 0.50 mol/l 0.0514 mol/l 100 = 90% 0.50 mol/l ( ) =.9683

7 Chapter 13 The number of half-lives needed to reduce the concentration of N O 5 to 10% can be calculated: 10 100 = 0.50n 0.10 = 0.50 n ln0.10 = nln0.50 n = 3.3 13.70. Collect and Organize For the decomposition reaction of PH 3 to P 4 and H, we are given that the reaction is first order with a rate constant, k, equal to 0.03 s 1 3 that reacts in 1 min if the initial partial pressure of PH 3 is 375 torr. The integrated rate law for a first-order reaction is ln[a] = kt + ln[a] 0 The partial pressure of PH 3 remaining after 1 min is ln[ph 3 ] = kt + ln[ph 3 ] 0 0.03 ln[ph 3 ] = s 60 s + ln ( 375 torr ) = 4.547 [PH 3 ] = e 4.547 = 94.3 torr The percentage of PH 3 that reacts in 1 min is 375 torr 94.3 torr 100 = 75% 375 torr The number of half-lives needed to reduce the concentration of N O 5 to 10% can be calculated: 5 100 = 0.50n 0.5 = 0.50 n ln0.5 = nln0.50 n =.0 13.71. Collect and Organize For the decomposition reaction of N O to N and O, we are given that the plot of ln[n O] versus time is linear. We are to write the rate law and then determine the number of half-lives it would take for the concentration of N O to become 6.5% of its original concentration. The integrated rate laws for zero-, first-, and second-order reactions with their half-life equations are as follows: [A] = kt + [A] 0 zero order t 1/ = [A] 0 /k ln[a] = kt + ln[a] 0 first order t 1/ = 0.693/k 1/[A] = kt + 1/[A] 0 second order t 1/ = 1/k[A] 0 (a) Since we are given that the plot of ln[n O] versus time is linear, the reaction is first order and the rate law is Rate = k[n O].

Chemical Kinetics 73 (b) The number of half-lives needed to reduce the concentration to 6.5% would be, where n = number of halflives, 6.5 n = 0.50 100 n 0.065 = 0.50 ln 0.065 = n ln 0.50 n = 4 The number of half-lives needed to reduce the concentration of a reactant to products is not dependent on the order of the reaction or on the magnitude of the rate constant. 13.7. Collect and Organize From the information given about the linearity of the plots of [C 4 H 6 ], ln[c 4 H 6 ], and 1/[C 4 H 6 ] versus time, we can determine the rate law for the dimerization of C 4 H 6 and then calculate the number of half-lives required for the concentration of C 4 H 6 to be reduced to 3.1% of its original concentration. The integrated rate laws for zero-, first-, and second-order reactions with their half-life equations are as follows: [A] = kt + [A] 0 zero order t 1/ = [A] 0 /k ln[a] = kt + ln[a] 0 first order t 1/ = 0.693/k 1/[A] = kt + 1/[A] 0 second order t 1/ = 1/k[A] 0 (a) Since we are given that the plot of 1/[C 4 H 6 ] versus time is linear, the reaction is second order and the rate law is Rate = k[c 4 H 6 ]. (b) The number of half-lives needed to reduce the concentration of C 4 H 6 to 3.1% would be, where n = number of half-lives, 3.1 n = 0.50 100 n 0.031 = 0.50 ln 0.031 = n ln 0.50 5.0 = n The number of half-lives needed to reduce the concentration of a reactant to a certain amount of product is not dependent on the order of the reaction or on the magnitude of the rate constant. 13.73. Collect and Organize From the data given for the concentration of 3 P over time, we are to determine the rate law and calculate the value for k for this radioactive decay. Radioactive decay follows first-order kinetics. This is confirmed in part b, where we are told to determine the first-order rate constant. A plot of ln[ 3 P] versus time for this decay gives a straight line with slope = k. The half-life of a first-order decay is given by 0.693 t1/ = k

74 Chapter 13 (a) Radioactivity (ln pci).4. 1.8 1.6 1.4 1. 1 y = 0.0485 +.303! R² = 1! 0 5 10 15 0 Time (d) The rate law for this radioactive decay is Rate = k[ 3 P]. (b) k = slope = 0.0485 day 1 0.693 (c) t 1/ = 1 = 14.3 days 0.0485 day The time needed for [ 3 P] to reduce to 1.00% of its original concentration, where n = number of half-lives, is 1.00 n = 0.50 100 n 0.0100 = 0.50 ln 0.0100 = n ln 0.50 n = 6.64 half-lives The number of days is n t1/ = 6.64 14.3 days = 95 days. 13.74. Collect and Organize From the data given for the second-order reaction that follows the concentration of HNO over time as it decomposes to NO, NO, and H O, we are to calculate the value of k and the half-life for this reaction. Because this reaction is second order in HNO, when we plot the concentration of HNO versus time according to the second-order integrated rate equation, 1/[A] = kt + 1/[A] 0 the slope of the line is equal to k and the half-life will be t 1/ = 1/k[A] 0. (a) The rate law for this second-order reaction is Rate = k[hno ]. A plot of the reciprocal of the concentration of HNO versus time for this second-order reaction gives the slope = k = 4.097 10 4 µm 1 min 1. 8 1/[HNO ] (µm 1 ) 7.5 7 6.5 6 y = 4.097E 04 + 6.410E+00 R² = 1.000E+00 0 1000 000 3000 4000 Time (min)

Chemical Kinetics 75 (b) The half-life of this reaction is 1 1 t 1/ = = k[a] 0 (4.097 10 4 µm 1 min 1 ) (0.1560 µm ) = 1.565 104 min, or 10.87 d The half-life of a second-order reaction decreases as the initial concentration of the reactant increases. For an initial concentration of 0.31 µm for this particular reaction, the half-life is 1 1 t 1/ = = k[a] 0 (4.097 10 4 µm 1 min 1 ) (0.310 µm ) = 7.83 103 min, or 5.433d 13.75. Collect and Organize Given that the dimerization of ClO to Cl O is second order, we are to determine the value of the rate constant and to calculate the half-life of this reaction. We are given a single data set and knowledge that the reaction is second order. For this second-order reaction a plot of 1/[ClO] versus time gives a straight line with a slope equal to k, the reaction rate constant. The half-life of this second-order reaction is 1 t1/ = k[clo] 0 where [ClO] 0 is the initial concentration of ClO used in the reaction. 1/[ClO], (molecules/cm 3 ) 1 3.00E-11.50E-11.00E-11 1.50E-11 1.00E-11 5.00E-1 0.00E+00 y = 5.398E 1 + 3.851E 1 R² = 1.000E+00 0 1 3 4 5 Time (s) The value of k for this reaction is k = slope = 5.40 10 1 cm 3 molecules 1 s 1, and the half-life of the reaction is as follows: t 1/ = 1 k[clo] 0 = 5.40 10 1 cm 3 molecules s 1.60 1011 molecules cm 3 = 0.71 s For a second-order reaction, as we increase the initial concentration of the reactant the half-life gets shorter. 13.76. Collect and Organize From the data given for the concentration of Cl O over time as it decomposes to ClO and the fact that the reaction is first order, we can calculate the rate constant and the half-life for the reaction.

76 Chapter 13 For a first-order reaction, the slope of the line of the linear plot of ln[a] versus time equals k. Once we have the value of k, the half-life may be calculated from 0.693 t1/ = k -16.4 ln[cl O ] (ln M) -16.6-16.8-17 -17. -17.4 y = 8.715E 04 1.653E+01 R² = 1.00E+00 0 00 400 600 800 1000 Time (µs) Because the slope = 8.7 10 4, k = 8.7 10 4 µs 1. The half-life of the reaction is 0.693 t 1/ = = 795 µs 8.7 10 4 1 µs From this information we could calculate how long it might take, for eample, for the concentration of Cl O to decrease to 15% of its original concentration. The integrated rate law for first-order reactions is ln[a] = kt + ln[a] 0 where, for this eample, [A] = 15, [A] 0 = 100, and k = 8.715 10 4 µs 1 : ln(15) = (8.715 10 4 µs 1 t) + ln(100) t =.18 10 3 µs 13.77. Collect and Organize From the data for the pseudo-first-order hydrolysis of sucrose provided, we are to write the rate law and determine the value of the pseudo-first-order rate constant, k. To obtain the value of k, we plot ln[sucrose] over time. The slope of the line is equal to k. The rate law is Rate = k[c 1 H O 11 ][H O] = k [C 1 H O 11 ]. -0.5 ln[sucrose] (ln M) -0.6-0.7-0.8 y = 5.190E 05 5.878E 01 R² = 9.76E-01-0.9 0 1000 000 3000 4000 5000 6000 Time (s) The pseudo-first-order plot gives k = slope = 5.19 10 5 s 1.

Chemical Kinetics 77 If we knew the concentration of water in the hydrolysis reaction, we could calculate the value of k by using k k = [HO] 0 13.78. Collect and Organize For the reaction of HO with O 3, we are asked to determine the pseudo-first-order rate constant given the eperimental data for [HO ] and [O 3 ] over time where the [O 3 ] is large and therefore remains constant throughout the reaction. Because we are given the actual [O 3 ], we can also calculate the second-order rate constant for the reaction. To determine the pseudo-first-order rate constant, we treat the data for [HO ] versus time as a first-order reaction (plot ln[ho ] versus t), which gives a straight line with slope = k' (the pseudo-first-order rate constant). The second-order rate constant can be calculated from the known eperimental concentration of ozone and the value of the pseudo-first-order rate constant: k' = k[o 3 ], or k = k'/[o 3 ] The first-order plot gives k' = slope = 1.03 10 ms 1. -1.6 ln[ho ] (ln M) -1.8-13 -13. y = 1.034E 0 1.65E+01 R² = 9.991E 01-13.4-13.6 0 0 40 60 80 100 Time (ms) The second-order rate constant, k, is k = 1.03 10 ms 1 1.0 10 3 M = 10.3 M 1 ms 1, or 1.0 10 4 M 1 s 1 When a second-order reaction is run so that one reactant is in large ecess, the reaction appears to obey firstorder kinetics for the other reactant. By this method, the rate constants of reactions involving more than one reactant can be eperimentally determined. 13.79. Collect and Organize We are asked to eplain how the magnitude of the activation energy for a reaction is related to the rate of the reaction. The activation energy is the minimum amount of energy required for colliding molecules to react. The greater the activation energy, the more energy required for the reaction to occur and the slower the reaction. This is because the higher the activation energy, fewer molecules will possess the minimum energy required for reaction. With fewer molecules reacting, the slower the reaction.

78 Chapter 13 A reaction with a lower activation energy is faster than one that is higher at the same temperature. 13.80. Collect and Organize We are to eplain whether all spontaneous reactions happen instantly at room temperature. In a spontaneous reaction, the free energy of the products is lower than the free energy of the reactants. In an instantaneous reaction the activation energy is near zero. No, just because a reaction is spontaneous (negative free energy), it does not mean that the reaction is instantaneous (or even fast). A spontaneous reaction can be either fast (low activation energy) or quite slow (large activation energy). A very fast, spontaneous reaction can be an eplosion. 13.81. Collect and Organize We are asked to describe the circumstances for a reaction in which the activation energy of the forward reaction is less than that of the reverse reaction. For the activation energy of the forward reaction to be less than that of the reverse reaction, the energy of the reactants must be closer to the energy of the transition state than the energy of the products. For the activation energy of the forward reaction to be less than the activation energy of the reverse reaction, the energy of the products must be lower than the energy of the reactants and therefore the reaction is spontaneous ( ΔG).

Chemical Kinetics 79 Just because a reaction is favored thermodynamically does not necessarily mean that it is fast. This reaction could have a quite high activation energy in the forward direction and therefore be quite slow. 13.8. Collect and Organize We are asked to describe the circumstances for a reaction in which the activation energy of the forward reaction is greater than that of the reverse reaction. For the activation energy of the forward reaction to be greater than that of the reverse reaction, the energy of the products must be closer to the energy of the transition state than the energy of the reactants. For the activation energy of the forward reaction to be greater than the activation energy of the reverse reaction, the energy of the reactants must be lower than the energy of the products and therefore the reaction is nonspontaneous (+ΔG). Just because a reaction is not favored thermodynamically under standard conditions does not necessarily mean that it does not proceed at all under nonstandard conditions. We will look at this more carefully when studying equilibrium and thermodynamics (Section 14.9). 13.83. Collect and Organize We are to eplain why the order of a reaction is independent of temperature, yet the value of k changes with temperature. We need to consider how temperature affects the motion and collision of the reactants. An increase in temperature increases the frequency and the kinetic energy at which the reactants collide. This speeds up the reaction, changing the value of k. The activation energy of the slowest step in the reaction, however, is not affected by a change in temperature and, therefore, the order of the reaction is unaffected. As a general 13.84. Collect and Organize For a reduction in the activation energy by ½, we are to assess whether that will increase the value of the rate constant by a factor of. The rate constant is related to the activation energy by the equation k = Ae E a /RT

80 Chapter 13 In the equation k = Ae E a /RT if A and T are held constant (R is constant as well), k will change by a factor of e 1/ when the activation energy is halved. The value of e 1/ is 0.6065. For the value of k to double, e 1/ would have to equal. Therefore, halving the activation energy would not double the value of k. The rate of the reaction is increased, as the value of the rate constant is reduced by a factor of approimately 0.61. In this problem we would not epect the value of the rate constant to double (meaning that the reaction is slower) if we reduce the activation energy. 13.85. Collect and Organize In comparing two first-order reactions with different activation energies, we are to decide which would show a larger increase in its rate as the reaction temperature is increased. We can use the Arrhenius equation to mathematically determine which reaction would be most accelerated by an increase in temperature: ln ln E a k = A RT Let s assume that T = T 1. For either reaction the difference in the rate constants is as follows: Ea Ea ln kt = ln A ln k 1 T = ln A RT1 RT Ea Ea Ea 1 1 ln kt ln k = + = T1 RT RT1 R T1 T But because T = T 1, Ea 1 1 Ea ln kt ln k T = = 1 R T1 T1 RT1 For E a = 150 kj/mol, 150 kj/mol ln kt ln k T = 1 RT1 For E a = 15 kj/mol, 15 kj/mol ln kt ln k T = 1 RT1 Comparing these as a ratio, 150 kj/mol ln kt ln k Tfor E 1 a = 150 kj/mol RT1 = = 10 ln k ln for 15 kj/mol T k T E 1 a = 15 kj/mol RT1 Therefore, the reaction with the larger activation energy (150 kj/mol) would be accelerated more than the reaction with the lower activation energy (15 kj/mol) when heated. Our derivation demonstrates that different reactions with different activation energies will not accelerate in the same way when they are heated.

Chemical Kinetics 81 13.86. Collect and Organize By looking at the Arrhenius equation, we can determine whether E a depends on temperature. The Arrhenius equation is ln ln E a k = A RT No. From the form of the equation, we see a dependence of k on temperature because the equation is in the form of y = m + b, where y = ln k and = 1/T. The rate constant, not E a, changes with temperature. The Arrhenius equation was actually first proposed by Jacobus van t Hoff, but Svante Arrhenius interpreted it and introduced the idea of an activation energy. 13.87. Collect and Organize We can use an Arrhenius plot of rate constant versus temperature for the reaction of O and O 3 to determine the activation energy (E a ) and the value of the frequency factor (A). The Arrhenius equation is Ea 1 ln k = + ln A R T If we plot ln k (y-ais) versus 1/T (-ais), we obtain a straight line with slope m = E a /R. The activation energy is therefore calculated from E a = slope R where R = 8.314 J/mol K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or e b = A. The Arrhenius plot gives a slope of 060 and a y-intercept of 0.00160. E a = ( 060 K 8.314 J/mol K) = 1.713 10 4 J/mol, or 17.1 kj/mol b = 0.00160 = ln A A = e 0.00160 = 1.00 Once we have the values of E a and A from the plot, we can calculate the value of k at any temperature. 13.88. Collect and Organize We can use an Arrhenius plot of rate constant versus temperature for the reaction of NO and O 3 to determine the activation energy (E a ) and to find the rate constant for the reaction at 300 K.

8 Chapter 13 The Arrhenius equation is Ea 1 ln k = + ln A R T If we plot ln k (y-ais) versus 1/T (-ais), we obtain a straight line with slope m = E a /R. The activation energy is therefore calculated from E a = slope R where R = 8.314 J/mol K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or e b = A. Once E a and A are known, we may use another form of the Arrhenius equation to calculate k at any temperature: Ea k = Aep RT The Arrhenius plot gives a slope of 450.6 and a y-intercept of 5.005. (a) E a = ( 450.6 K 8.314 J/mol K) =.04 10 4 J/mol, or 0.4 kj/mol (b) To calculate k at 300 K we first need the value of A: A = e 5.005 = 7.37 10 10 The value of k at 300 K is 4.04 10 J/mol k = 7.37 10 ep =.03 10 M s 8.314 J/mol K 300 K 10 7 1 1 Alternatively, k can be calculated from the original Arrhenius equation: 4 Ea.04 10 J/mol ln k = + ln A= + 5.005 = 16.83 RT 8.314 J/mol K 300 K 7 1 1 k =.04 10 M s 13.89. Collect and Organize We can use an Arrhenius plot of rate constant versus temperature for the reaction of N with O to form NO to determine the activation energy (E a ), the frequency factor (A), and the rate constant for the reaction at 300 K. The Arrhenius equation is Ea 1 ln k = + ln A R T If we plot ln k (y-ais) versus 1/T (-ais), we obtain a straight line with slope m = E a /R. The activation energy is therefore calculated from E a = slope R

Chemical Kinetics 83 where R = 8.314 J/mol K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or e b = A. Once E a and A are known, we may use another form of the Arrhenius equation to calculate k at any temperature: k Aep E a = RT The Arrhenius plot gives a slope of 37758 and a y-intercept of 4.641. (a) E a = ( 37758 K 8.314 J/mol K) = 3.14 10 5 J/mol, or 314 kj/mol (b) A = e 4.641 = 5.03 10 10 5 10 3.14 10 J/mol 44 1/ 1 (c) k = 5.03 10 ep = 1.06 10 M s 8.314 J/mol K 300 K Alternatively, k can be calculated from the original Arrhenius equation: 5 Ea 3.14 10 J/mol ln k = + ln A= + 4.641 RT 8.314 J/mol K 300 K = 101.5 101.5 44 1/ 1 k = e = 1.07 10 M s 13.90. Collect and Organize We can use an Arrhenius plot of rate constant versus temperature for the decomposition of N O 5 to determine the activation energy (E a ) and the rate constant for the reaction at 300 K. The Arrhenius equation is Ea 1 ln k = + ln A R T If we plot ln k (y-ais) versus 1/T (-ais), we obtain a straight line with slope m = E a /R. The activation energy is therefore calculated from E a = slope R where R = 8.314 J/mol K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or e b = A. Once E a and A are known, we may use another form of the Arrhenius equation to calculate k at any temperature: k Aep E a = RT The Arrhenius plot gives a slope of 11781 and a y-intercept of 30.164.

84 Chapter 13 (a) E a = ( 11781 K 8.314 J/mol K) = 9.79 10 4 J/mol, or 97.9 kj/mol (b) To calculate k at 300 K we first need the value of A A = e 30.164 = 1.6 10 13 The value of k at 300 K is 4 9.79 10 J/mol k = 1.6 10 ep = 1.13 10 s 8.314 J/mol K 300 K This reaction will be relatively slow, as k < 1. 13 4 1 13.91. Collect and Organize We can use an Arrhenius plot of rate constant versus temperature for the reaction of ClO and O 3 to determine the activation energy (E a ) and the value of the frequency factor (A). The Arrhenius equation is Ea 1 ln k = + ln A R T If we plot ln k (y-ais) versus 1/T (-ais), we obtain a straight line with slope m = E a /R. The activation energy is therefore calculated from E a = slope R where R = 8.314 J/mol K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or e b = A. The Arrhenius plot gives a slope of 4698.7 and a y-intercept of 7.87. E a = ( 4698.7 K 8.314 J/mol K) = 3.91 10 4 J/mol, or 39.1 kj/mol b = 7.87 = ln A A = e 7.87 = 1.7 10 1

Chemical Kinetics 85 Once we have the values of E a and A from the plot, we can calculate the value of k at any temperature. 13.9. Collect and Organize Given the value of the rate constant for the reaction of Cl with CH 4 to form HCl and CH 3 at 98 K and at three additional temperatures, we can use an Arrhenius plot to calculate the activation energy (E a ) and the frequency factor (A) for the reaction. The Arrhenius equation is Ea 1 ln k = + ln A R T If we plot ln k (y-ais) versus 1/T (-ais), we obtain a straight line with slope m = E a /R. The activation energy is therefore calculated from E a = slope R where R = 8.314 J/mol K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or e b = A. The Arrhenius plot gives a slope of 1387.5 and a y-intercept of.567. E a = ( 1387.5 K 8.314 J/mol K) = 1.15 10 4 J/mol, or 11.5 kj/mol A = e.567 = 6.3 10 9 The rate constant for this reaction is fairly large, so we epect that the reaction will proceed quickly to products. 13.93. Collect and Organize, we are to mperature, the rate of a reaction doubles. We can use the form of the Arrhenius equation relating the rate constants for a reaction at different temperatures: ln k 1 = E a k R 1 1 T T 1 For the equation we use k =.00 10 8 s 1, T T 1 E a =.14 10 4 J/mol with R = 8.314 J/mol K.

86 Chapter 13 ln The ratio of these rate constants is k 1.00 10 8 s =.14 104 J/mol 1 8.314 J/mol K k 1.00 10 8 s 1 = e0.03391 = 1.034 k 1 = 1.034.00 10 8 s 1 k 610 C 1 873 K 1 883 K = 0.03391 ( ) =.07 10 8 s 1 =.07 108 s 1 = 1.04 k 600 C.00 10 8 1 s does not apply to this temperature domain. This rule applies best when the reaction temperatures are near 98K. This problem illustrates well how general rules do not apply to every situation, so be cautious in using general rules to make predictions. 13.94. Collect and Organize For the decomposition of CH 3 CHF to CH CHF and HF we are to determine to what temperature we would k is 5.8 10 6 s 1 and the activation energy is 65 kj/mol. We can use the form of the Arrhenius equation relating the rate constants for a reaction at different temperatures to solve this problem. ln k 1 = E a k R 1 1 T T 1 For the equation we can use the ratio of k 1 /k = 4 with T E a =.65 10 5 J/mol with R = 8.314 J/mol K. ln 4 1 =.65 105 J/mol 8.314 J/mol K 1 733 K 1 T 1 8.314 J/mol K ln4.65 10 5 J/mol = 1 733 K 1 T 1 = 4.3493 10 5 K 1 1 4.3493 10 5 K 1 733 K = 1 = 1.307 10 3 K 1 T 1 T 1 = 757 K This is only a increase in temperature to make the reaction go four times as fast. 13.95. Collect and Organize By comparing the rate laws for two reactions, we can determine whether their mechanisms are similar. For the reaction between NO and H, For the reaction between NO and Cl, Rate = k[no] [H ] Rate = k[no][cl ]

Chemical Kinetics 87 No. The different rate laws for the two reactions indicate different mechanisms. However, even if two rate laws were similar, this would not necessarily mean that their reaction mechanisms are similar. 13.96. Collect and Organize For two reactions, one of NO with Cl and the other of NO with F, we consider whether their identical rate laws mean that they might have similar mechanisms. No difference eists in the forms of the two rate laws given. Both are first order in the two reactants, NO or NO and X, and both are second order overall. Yes, when the rate laws are the same, the reactions can have similar mechanisms. Be careful here. Reactions with the same form of the rate laws may proceed by very different mechanisms. 13.97. Collect and Organize / We are to identify the conditions under which a bimolecular reaction shows pseudo-first-order behavior. Pseudo-first-order kinetics occurs when one reactant is in sufficiently high concentration that it does not change appreciably during the reaction. We solved problems relating to pseudo-first-order reactions earlier in this chapter (Problems 13.77 and 13.78). 13.98. Collect and Organize For a reaction that is zero order in a reactant, we are to determine whether that means the reactant is not involved in any collisions with other reactants during the reaction. When a reactant is zero order it does not appear in the overall rate equation. No. The reactant may be used and therefore must collide with other reactants, but it is not involved in the ratedetermining step of the reaction. The rate law epresses only the concentration dependence of those reactants (and intermediates) involved in the slowest (rate determining) step of the mechanism. 13.99. Collect and Organize We are asked to draw reaction profiles that fit the reaction A B (which has E a = 50.0 kj/mol) for three different mechanisms: (a) a single elementary step, (b) a two-step reaction with E a,step = 15 kj/mol, and (c) a two-step reaction in which the second step is rate determining. The reaction profile plots energy versus reaction progress and shows the presence of an activated comple at highest energy, the reaction intermediates in the valleys between reactant and products, and the relative activation energies in multistep reactions. The rate-determining step is the slowest step in the mechanism and has the largest activation energy in the reaction profile.

88 Chapter 13 These reaction profiles could also be drawn for a nonspontaneous reaction. Reactant A would then be lower in energy than product B. 13.100. Collect and Organize We are asked to draw reaction profiles that fit the reaction A + B C D + E for three different mechanisms. The reaction profile plots energy versus reaction progress and shows the presence of an activated comple at highest energy, the reaction intermediates in the valleys between reactant and products, and the relative activation energies in multistep reactions. The rate-determining step is the slowest step in the mechanism and has the largest activation energy in the reaction profile. (a) When C is an activated comple: (b) When the first step is rate determining and C is an intermediate:

Chemical Kinetics 89 (c) When the second step is rate determining and C is an intermediate: These reaction profiles could also be drawn for a nonspontaneous reaction. For eample, when the first step of a two-step reaction is rate determining with C as the intermediate, the reaction profile would look like this: 13.101. Collect and Organize For each elementary step given, we are to write the rate law and determine whether the step is uni-, bi-, or termolecular. The rate law for an elementary step in a mechanism is written in the form Rate = k[a] [B] y [C] z where A, B, and C are the reactants involved in the elementary reaction and, y, and z are the stoichiometric coefficients for the respective reactants in the elementary reaction. (a) Rate = k[so Cl ]. Because this elementary step involves only a molecule of SO Cl, it is unimolecular. (b) Rate = k[no ][CO]. Because this elementary step involves a molecule of NO and a molecule of CO, it is bimolecular. (c) Rate = k[no ]. Because this elementary step involves two molecules of NO, it is bimolecular. Termolecular elementary reactions are rare. 13.10. Collect and Organize For each elementary step given, we are to write the rate law and determine whether the step is uni-, bi-, or termolecular. The rate law for an elementary step in a mechanism is written in the form Rate = k[a] [B] y [C] z where A, B, and C are the reactants involved in the elementary reaction and, y, and z are the stoichiometric coefficients for the respective reactants in the elementary reaction.

90 Chapter 13 (a) Rate = k[cl][o 3 ]. Because this elementary step involves a Cl atom and an O 3 molecule, it is bimolecular. (b) Rate = k[no ]. Because this elementary step involves two NO molecules, it is bimolecular. (c) Rate = k[ 14 C]. Because this elementary step involves only a 14 C atom, it is unimolecular. Termolecular elementary reactions are rare. 13.103. Collect and Organize From three elementary steps that describe a reaction mechanism, we are to write the overall chemical equation. To write the overall chemical reaction we need to add the elementary steps, being sure to cancel the intermediates in the reaction. N O ( g) NO ( g) + NO ( g) NO ( ) NO ( g) + O( g) 5 3 3 g g g O( ) O ( ) N O ( g) + O( g) NO ( g) + O ( g) 5 In this reaction, NO 3 is a reaction intermediate. It is generated in the reaction but consumed in a subsequent step in the mechanism. 13.104. Collect and Organize From three elementary steps that describe a reaction mechanism, we are to write the overall chemical equation. To write the overall chemical reaction we need to add the elementary steps, being sure to cancel the intermediates in the reaction. ClO (aq) + H O( ) HClO(aq) + OH (aq) I (aq) + HClO(aq) HIO(aq) + Cl (aq) OH (aq) + HIO(aq) H O( ) + IO (aq) ClO (aq) + I (aq) Cl (aq) + IO (aq) In this reaction HClO, OH, and HIO are reaction intermediates. They are generated in the reaction but consumed in subsequent steps in the mechanism. 13.105. Collect and Organize We are given the mechanism by which N reacts with O to form NO. For a given rate law of Rate = k[n ][O ] 1/ we are to determine which step in the mechanism is the rate-determining step. To determine which step in the proposed mechanism might be the slowest, we can write the rate law for the mechanism when the first, second, or third step is slow and then match the theoretical rate law to the eperimental rate law.

Chemical Kinetics 91 If the first step is slow, the rate law is Rate = k 1 [O ] This does not match the eperimental rate law, so the first step is not the slowest step in the mechanism. If the second step is slow, the rate law is Rate = k [O][N ] Because O is an intermediate, we use the first step to epress its concentration in terms of concentrations of the reactants. For a fast step occurring before a slow step in a mechanism, Rate forward = Rate reverse k 1 [O ] = k 1 [O] Rearranging to solve for [O], 1/ k 1 [O] = [ O ] k 1 Substituting this into the rate law from the second step gives k Rate = k 1 O k 1 1/ 1/ N = k O N This rate law matches the eperimental rate law, so the rate-determining step is the second step. We should check to see whether the mechanism of the third step is the slow step and might also give the eperimental rate law. From the logic above, Rate = k 3 [N][O] From the second fast step in the mechanism, k [O][N ] = k [NO][N] Solving for [N], an intermediate, gives k[ O][ N] [N] = k NO [ ] From the first fast step in the mechanism, k 1 [O ] = k 1 [O] solving for [O] gives k1 [O] = [ O ] k 1 Substituting these epressions into the rate law from the third step in the mechanism gives k Rate = k k 1 [O ][N ] 3 k k 1 [NO] = k[n ][O ] [NO] This rate law does not match the eperimental rate law. Just because the rate law for a mechanism matches the eperimental rate law does not mean that the mechanism is the correct mechanism. Another mechanism might also give the same eperimental rate law. 13.106. Collect and Organize Given a proposed mechanism for the decomposition of H O, we are to determine which step is the ratedetermining (slow) step if the reaction is known by eperiment to be first order in H O. The slowest step in a mechanism determines the rate law. From each elementary reaction we can write a rate law epression.

9 Chapter 13 If the first step is the rate-determining step, the rate law is Rate = k 1 [H O ] This is consistent with the reaction being first order in the concentration of H O, so the first step is likely to be the rate-determining step. If the second step were rate determining, the rate law would be Rate = k [H O ][OH] This is also first order in H O but is also first order in OH concentration. OH in this reaction is a reaction intermediate. Because the first step is fast and reversible, we can write k f [H O ] = k r [OH] Rearranging this to solve for the concentration of OH, the reaction intermediate, gives 1 k f [OH] = [HO ] kr Substituting this concentration of OH into the rate law epression for the rate-determining step gives Rate = k [H O ] k 1 1 k f [H k O ] = k f [H O ] 3/ r Thus, the rate law derived from the mechanism where the second step is rate determining is inconsistent with the eperimentally observed rate law. 13.107. Collect and Organize We are given the mechanism by which NO reacts with Cl to produce NOCl. For a given rate law of Rate = k[no][cl ] we are to determine which step in the mechanism is the rate-determining step. To determine which step in the proposed mechanism might be the slowest, we can write the rate law for the mechanism when the first, second, or third step is slow and then match the rate law to the eperimental rate law. If the first step is slow, the rate law is Rate = k 1 [NO][Cl ] This matches the eperimental rate law, so the first step is the rate-determining step. We should check to see whether the rate law for the mechanism with the second step as the slow step, Rate = k [NOCl ][NO] might also give the eperimental rate law. If the second step is slow, then Rate 1 = Rate 1 k 1 [NO][Cl ] = k 1 [NOCl ] k1 [NOCl ] = [ NO][ Cl] k 1 Substituting this into the rate law epression gives k Rate = k 1 NO k Cl 1 NO = k NO Cl This does not match the eperimental rate law, so the second step in the mechanism is not the rate-determining step. Just because the rate law for a mechanism matches the eperimental rate law does not mean the mechanism is the correct mechanism. Another possible mechanism might also give the same eperimental rate law. k r

Chemical Kinetics 93 13.108. Collect and Organize Given a proposed mechanism for the thermal destruction of ozone, we are to consider the relative rates and reversibility of the elementary steps to determine what properties are consistent with the eperimental observation that the reaction is second order in O 3. The rate law is determined by the slowest step in the mechanism. We can write the rate epressions for the two possibilities the first step as slow or the second step as slow and then consider the consistency of those two possible rate laws with the observation that the reaction is second order in ozone. If the first step is slow, Rate = k[o 3 ] This is only first order in O 3, so the first step as the rate-determining step in the mechanism is inconsistent with the eperimental observation for this reaction. If the second step is slow, then the first step is fast and reversible and the rate of the forward reaction is equal to the rate of the reverse reaction: k f [O 3 ] = k r [O][O ] When the second step is rate determining, the rate law epression will be Rate = k [O][O 3 ] Combining these equations by replacing [O] in the rate law epression for the second slow step with kf[o 3] [O] = kr[o ] obtained from the reversible first step of the mechanism gives Rate = k k [O f 3 ] k r [O ] This rate law is consistent with the observation that the reaction is second order in O 3. The rate law for this reaction also shows an inverse dependence of the reaction rate on the concentration of O. This means that if the concentration of O increases, the reaction slows down. 13.109. Collect and Organize From the mechanisms given, we are to determine which are possible for the thermal decomposition and which are possible for the photochemical decomposition of NO. We are given the rate laws: for the thermal decomposition reaction, Rate = k[no ] ; for the photochemical decomposition, Rate = k[no ]. Using the slowest elementary step in the mechanism, we can write the rate law epression for each mechanism and then determine which is consistent with the order of the reaction given for each process. For mechanism a, the first step in the mechanism is slow, so the rate law is Rate = k[no ] For mechanism b, the second step in the mechanism is slow, so the rate law is Rate = k [N O 4 ] Using the first step to epress [N O 4 ] in terms of the concentrations of the reactant NO gives k 1 [NO ] = k 1 [N O 4 ] k1 [N [ ] O 4] = NO k 1

94 Chapter 13 Substituting into the rate epression from the second step, Rate = k k 1 NO k = k NO 1 For mechanism c, the first step in the mechanism is slow, so the rate law is Rate = k[no ] Therefore, mechanisms b and c are consistent with the thermal decomposition of NO and mechanism a is consistent with the photochemical decomposition of NO. To distinguish between the two possible mechanisms for thermal decomposition, we might try to detect the different intermediates formed in each. Detection of the formation of N O 4 would support mechanism b over mechanism c. 13.110. Collect and Organize From the possible mechanisms, we are to determine which occurs for the thermal decomposition and which for the photochemical decomposition of NO. We are given that the thermal decomposition reaction is second order in NO and that the photochemical decomposition is first order in NO. Using the slowest elementary step in the mechanism, we can write the rate law epression for each mechanism and then determine which is consistent with the order of the reaction given for each process. Because the first step in each mechanism is the rate-determining step, the rate laws are as follows: (a) Rate = k[no ] (b) Rate = k[no ] (c) Rate = k[no ] Therefore, mechanisms a and b are consistent with the thermal decomposition process and mechanism c is consistent with the photochemical decomposition process. To distinguish between the two possible mechanisms for thermal decomposition, we might try to detect the different intermediates formed in each. The detection of N O 4, N O 3, or N O would support mechanism a, whereas the detection of NO 3 would support mechanism b. 13.111. Collect and Organize We are asked whether a catalyst affects both the rate and the rate constant of a reaction. A catalyst speeds up a reaction by providing an alternate pathway (mechanism) to the products; this alternate pathway has a lower activation energy. Yes. Because the reaction is faster (affecting the rate) and the activation energy is lowered (affecting the value of k), a catalyst affects both the rate of the reaction and the value of the rate constant. A negative catalyst that slows down a reaction would increase E a and decrease k for a reaction. We call these negative catalysts inhibitors. 13.11. Collect and Organize We are asked whether the rate law for a catalyzed reaction is the same as that for the uncatalyzed reaction. A catalyst is used in the reaction but later is regenerated.

Chemical Kinetics 95 The rate law for a catalyzed reaction, like any other reaction, depends on the species involved in the ratelimiting step. The rate law for a catalyzed reaction may be the same as the rate law for an uncatalyzed reaction, but if, for eample, the catalyst is involved in the rate-limiting step, its concentration appears in the rate law. This pathway, then, would have a different rate law from that of the uncatalyzed reaction. Remember that catalysts enhance the rate of a reaction by providing a different, lower-activation-energy pathway to the products. 13.113. Collect and Organize We are asked whether a substance (catalyst) that increases the rate of a reaction also increases the rate of the reverse reaction. A catalyst speeds up a reaction by providing an alternate, lower-activation-energy pathway (mechanism) to the products. Yes, both the reverse and forward reaction rates are increased when a catalyst is added to a reaction. The activation energies of both processes are lowered by the different pathway that the catalyst provides for the reaction. Likewise, an inhibitor would decrease the rates of both forward and reverse reactions. 13.114. Collect and Organize We are asked whether CO is a catalyst in the reaction between NO and CO because it does not appear in the rate law. A catalyst is used in the reaction but later is regenerated, and a catalyst speeds up a reaction by providing an alternate, lower-activation-energy pathway (mechanism) to the products. No. To function as a catalyst, the CO would have to speed up the reaction and not be consumed. If it does not lower the E a for the reaction and later is generated in the reaction, it is not a catalyst. The CO may not appear in the rate law because it may not be involved in the rate-limiting step of the mechanism.

96 Chapter 13 13.115. Collect and Organize We are to eplain why the concentration of a homogeneous catalyst does not appear in the rate law. A catalyst is used in a reaction and later regenerated. The concentration of a homogeneous catalyst may not appear in the rate law because the catalyst itself is not involved in the rate-limiting step. If the catalyst is involved in the slowest step of the mechanism, however, it is involved in the rate law. 13.116. Collect and Organize We are to identify the better way to determine the rate constant for a slow reaction: adding a catalyst or increasing the temperature. A catalyst provides a lower activation energy to the reaction through a different mechanism to speed up a reaction. Heating a reaction causes the reactants to collide more frequently and more effectively without changing the mechanism. Raising the temperature to determine the rate constant for a slow reaction would be better because the mechanism for the reaction is not changed. Using the form of Arrhenius s equation in which we compare two rate constants at two temperatures, Ea 1 ln k1 = + ln A R T1 Ea 1 ln k = + ln A R T k Ea 1 1 ln k ln k1 = ln = k1 R T1 T we could calculate k 1 at the lower temperature T 1, having measured k at the higher temperature T, and knowing E a. 13.117. Collect and Organize Given the mechanism for the reaction of NO and N O to form N and O, we are to determine whether NO or N O is used in the reaction as the catalyst. A catalyst is used in a reaction and later regenerated and provides a lower-energy pathway to the products by lowering the activation energy of the reaction, thereby speeding up the reaction. We can assume that the presence of either NO or N O, if either is a catalyst, increases the rate of the reaction. In eamining the mechanism we see that N O is a reactant, not a catalyst, but that NO is a catalyst because it is used in the reaction and then regenerated. Thus, NO is a catalyst for the decomposition of N O. If the slow step of this mechanism were the first step, the rate law would be Rate = k[no][n O]

Chemical Kinetics 97 If the second step were slow, the rate law would be Rate = k NO N O N 13.118. Collect and Organize Given the mechanism for the destruction of O 3 in the presence of NO, we are to eplain why NO is functioning as a catalyst. A catalyst is used in a reaction and later regenerated. It provides a lower-energy pathway to the products by lowering the activation energy of the reaction, thereby speeding up the reaction. NO is used in the reaction in step 1 and regenerated in step. Thus, as long as this mechanism has a lower energy pathway to form O from O and O 3, NO is functioning as a catalyst. If, however, the presence of NO slows down the reaction, it would be an inhibitor. 13.119. Collect and Organize Using the Arrhenius equation, we can compute and compare the rate constants for the reaction of O 3 with O versus the reaction of O 3 with Cl. We are given values A and E a for each reaction at 98 K. The Arrhenius equation is Ea ln k = + ln A RT For the reaction of O 3 with O, 17.1 10 3 J/mol ln k = 8.314 J/mol K 98 K + ln 8.0 10 1 cm 3 /molecules s ln k = 3.45 k = 8.05 10 15 cm 3 /molecules s For the reaction of O 3 with Cl,.16 10 3 J/mol ln k = 8.314 J/mol K 98 K + ln.9 10 11 cm 3 /molecules s ln k = 5.14 k = 1.1 10 11 cm 3 /molecules s Therefore, the reaction of O 3 with Cl has the larger rate constant. ( ) ( ) Our answer is consistent with a qualitative look at the activation energies and frequency factors for the two reactions. The higher activation energy and lower frequency factor for the reaction of O 3 with O give a smaller reaction rate constant. 13.10. Collect and Organize Using the Arrhenius equation, we can compute and compare the rate constants for the reaction of O 3 with Cl versus the reaction of O 3 with NO.

98 Chapter 13 We are given values of A and E a for each reaction at 98 K. The Arrhenius equation is Ea ln k = + ln A RT For the reaction of O 3 with Cl,.16 10 3 J/mol ln k = 8.314 J/mol K 98 K + ln(.9 10 11 cm 3 /molecules s) ln k = 5.14 k = 1.1 10 11 cm 3 /molecules s For the reaction of O 3 with NO, 11.6 10 3 J/mol ln k = 8.314 J/mol K 98 K + ln(.0 10 1 cm 3 /molecules s) ln k = 31.6 k = 1.85 10 14 cm 3 /molecules s The reaction with the larger rate constant is the reaction of O 3 with Cl. Our answer is consistent with a qualitative look at the activation energies and frequency factors for the two reactions. The higher activation energy and lower frequency factor for the reaction of O 3 with NO give a smaller reaction rate constant. 13.11. Collect and Organize We are to eplain why a glowing wood splint burns faster in a test tube filled with O than in air. Air is composed of about 1% O. When the concentration of a reactant (O for the combustion reaction) increases, the rate of reaction also increases. As we place the glowing wood in pure O, the rate of combustion increases. If the wood splint were placed in a test tube filled with argon, the combustion reaction would stop. 13.1. Collect and Organize We are to eplain why a spark is required to ignite the spontaneous combustion of propane in a barbecue grill. The spark adds energy to the propane air (oygen) miture. The spark provides the energy to the propane air miture needed to overcome the activation energy barrier to start the combustion reaction. Once started, the combustion reaction itself is self-sustaining. 13.13. Collect and Organize We are to eplain why a person submerged in cold water is less likely to have a lack of oygen for a given period than a person submerged in a warm pool.

Chemical Kinetics 99 Chemical reactions are slower at colder temperatures than at warmer temperatures. The bodily reactions that use O are slower at colder temperatures, so the person submerged in an ice-covered lake uses less of the already dissolved oygen in his or her system than the person in a warm pool. Rapid-cooling technology is being investigated at Argonne National Laboratory for use in surgery patients and heart attack victims to reduce the damage done to cells by lack of oygen in the blood. 13.14. Collect and Organize We are asked to eplain why when we have a quadrupling of a reaction rate that it may not correspond to a reaction rate order of 4. Rate epressions show the relationship between the rate of a reaction and the concentration of the reactants raised to an eponent. Rate A To quadruple the rate of a reaction means that rate = 4 rate 1. To have a reactant be fourth order means that the concentration term in the rate equation will be raised to the fourth power. For eample, if Rate 1 NO 4, where NO = 1 M Rate 16 rate 1 if NO is doubled to M If a reaction is second order, as in the case of then if [NO] is doubled from 1 M to M Rate 1 NO Rate = 4 Rate 1 13.15. Collect and Organize In the case where rate reverse << rate forward, we are to consider whether the method to determine the rate law (initial concentrations and initial rates) would work at other times, not just at the start of the reaction. If so, we are to specify which concentrations might be used to determine the rate law. The method that uses initial rates and concentrations to determine the rate law is under the condition in which no reverse reaction is occurring. Yes, we could use this method at other times, not just t = 0, to determine the rate law if the rate of the reverse reaction is much slower than the forward reaction as long as [products] << [reactants] at the time so that no appreciable reverse reaction is occurring. We will see in Chapter 15 that when the rate of the reverse reaction equals the rate of the forward reaction, the reaction is at equilibrium.

100 Chapter 13 13.16. Collect and Organize Given a statement relating reaction rate and the rate constant directly to the number of collisions occurring in a reaction and the concentrations of the reactants, we are to determine what is incorrect about the statement. The rate law epression for a reaction describes the relationship between the concentration of reactants in the reaction and the rate of the reaction. The rate constant in the rate law (k) relates the rate of the reaction to the concentration of the reactants in a chemical reaction. In this statement, it is true that the reaction rate depends on the concentration of the reactants as stated eplicitly by the rate law epression. The reaction rate is also dependent on the number of collisions of the reactants, as more frequent collisions (which can be accomplished by increasing the reactant concentrations) will increase the reaction rate. However, the rate constant is not dependent on either the concentrations or the number of collisions of the reactants. The value of k is unique to a reaction and changes only with temperature or the use of a catalyst. The Arrhenius equation k = Ae E a RT shows that k is dependent on the frequency factory, A; the activation energy, E a ; and the temperature, T. 13.17. Collect and Organize In the plot of 1/[X] 1/[X] 0 as a function of time, t, we are asked how the rate constant, k, can be determined. The plot of 1/[X] 1/[X] 0 as a function of time, t, is the plot for a second-order rate equation. Rate = k X 1 = kt + 1 X X 0 In this plot 1/[X] 1/[X] 0 divided by t t 0 is the slope of the line that corresponds to k, the reaction rate constant. All we need to do to determine k from this plot is to determine the slope of the line. The integrated form of the rate law allows us to obtain the value of k from the concentration-versus-time data from a single eperiment. 13.18. Collect and Organize We are asked to eplain why first-order reactions are common, second-order reactions are less common, and third-order reactions are rare. First-order reactions involve a single reactant molecule only. Second-order reactions involve two molecules colliding. Third-order reactions involve three molecules colliding simultaneously. For a first-order reaction to occur, the single reactant molecule needs to have only sufficient energy to react to become products. It does not depend directly on collisions with other molecules, so first-order reactions are quite common. For a second-order reaction to occur, two reactant molecules must collide with sufficient energy and in the correct orientation to produce products. This is a higher requirement for a reaction, so second-order reactions are less common than first-order reactions. Third-order reactions are very rare because they require

Chemical Kinetics 101 that three molecules collide simultaneously with sufficient energy and in the correct orientation for the reaction to occur. Termolecular reactions are often eplained by two consecutive reactions, such as where once two species have reacted to form a highly energetic intermediate (AB*), an inert chemical species (M) collides with the intermediate. Ecess energy from the intermediate is then transferred to M. A + B AB* AB* + M C + M 13.19. Collect and Organize We are asked to eplain why an elementary step may not have a rate law that is zero order. An elementary step in a reaction mechanism describes the collisions of molecular or atomic species taking place in a reaction. For an elementary step to take place, some involvement from a molecular or atomic species must occur. Therefore, there cannot be no dependence (or zero order) of the reactant in an elementary step of a reaction mechanism. Because most elementary steps are either unimolecular or bimolecular, the rate epressions of elementary steps are usually first or second order. 13.130. Collect and Organize Given the balanced equation for the decomposition of N O 5, we are asked to relate the rate of change in [N O 5 ] to that of [NO ] and [O ]. From the balanced equation, the rate of formation of products and consumption of reactants is Rate = 1 Δ N O 5 = 1 4 Δ NO = Δ O The rate of consumption of N O 5 is one-half the rate of formation of NO and twice the rate of formation of O. Generally, in measuring reaction kinetics we monitor the rate of consumption of a reactant to obtain the initial rate of the reaction. 13.131. Collect and Organize Given the balanced equation for the reaction between NO and O 3 to produce N O 5 and O, we are asked to relate the rates of change in [NO ], [O 3 ], [N O 5 ], and [O ]. From the balanced equation, the rate of formation of products and consumption of reactants is Rate = 1 Δ NO = Δ O 3 = Δ N O 5 = Δ O The rate of consumption of O 3 is the same as the rate of formation of N O 5 and O and one-half the rate of consumption of NO.

10 Chapter 13 The rate of consumption of N O 5 is half the rate of formation of NO and twice the rate of formation of O. 13.13. Collect and Organize Given data for two eperiments in which the initial rate of the decomposition of N O 5 was measured at different concentrations of N O 5, we are to determine the order of the reaction. We can determine the order of the reaction with respect to N O 5 by comparing the rate of the two reactions on changing the concentration: rate [N O 5] 0, = rate 1 [NO 5] 0,1 This reaction is first order. 3.6 10 5 M/s 1.8 10 5 M/s = 0.100 M 0.050 M.0 =.0 = 1 Not all decomposition reactions are first order. The decomposing species might have to collide with another molecule of itself in the rate-determining step of the reaction, in which case the reaction is second order in the decomposing reactant. 13.133. Collect and Organize We can write the rate law from the order of the decomposition reaction determined in Problem 13.13. From that we are to calculate the value of the rate constant at the eperimental temperature and write the complete rate law epression. From Problem 13.13, we know that the reaction is first order in [N O 5 ]. Because this reaction is a first-order decomposition reaction, the rate law epression is Rate = k[n O 5 ] Using the data in eperiment 1 in Problem 13.13, 1.8 10 5 M/s = k 0.050 M k = 3.6 10 4 s 1 The complete rate law epression is then Rate = (3.6 10 4 s 1 ) [N O 5 ] The data from eperiment in Problem 13.13 would give the same value of k. 13.134. Collect and Organize In the reaction of NO with Cl, the rate of reaction was measured for various concentrations of both reactants. From the data we can determine the rate law and calculate the rate constant k, which will allow us to determine the initial rate of reaction in eperiment 4. To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two eperiments in which the concentration of that reactant changes but the

Chemical Kinetics 103 concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law epression. To calculate the rate constant for the reaction, we rearrange the rate law to solve for k and use the data from any of the eperiments. Finally, we can calculate the rate of the reaction for two different concentrations of the reactants by substituting those concentrations and the calculated value of k into the rate law epression. Using eperiments 1 and, we find that the order of the reaction with respect to Cl is : rate [Cl ] 0, = rate 1 [Cl ] 0,1 5.70 M/s 0.30 M = 0.63 M/s 0.10 M 9.0 = 3.0 = Using eperiments 1 and 3, we find that the order of the reaction with respect to NO is 1: rate [NO] 3 0,3 = rate 1 [NO] 0,1.58 M/s 0.80 M = 0.63 M/s 0.0 M 4.1 = 4.0 = 1 This gives a rate law for the reaction of Rate = k[no][cl ] Using eperiment 1, we can calculate k: M 0.63 = k ( 0.0 M ) ( 0.10 M ) s 1 k = 315 M s Using the calculated value of k and the [NO] 0 and [Cl ] 0, we calculate the rate for eperiment 4: Rate = 315 M s 1 0.40 M (0.0 M) = 5.0 M/s We may use any of the eperiments in the table to calculate k. Each eperiment s data give the same value of k as long as the eperiments were all run at the same temperature. 13.135. Collect and Organize For the reaction of NO with O 3 to produce NO and O we can use the information that the reaction is first order in both NO and O 3 along with the values of the rate constants at two different temperatures to determine whether the reaction occurs in a single or many steps and to calculate the activation energy, the rate of the reaction at another concentration of the reactants, and the rate constants at two other temperatures. To answer the questions we need to use the Arrhenius equation, Ea ln k = + ln A RT and the rate law epression, which states that the reaction is first order in both NO and O 3 : Rate = k[no][o 3 ]

104 Chapter 13 (a) Because the rate law in which the reaction is first order in both NO and O 3 is consistent with that in which the reaction would occur in a single step, this reaction might indeed occur in a single step. (b) We can calculate the activation energy for the reaction by comparing the rate constant at the two temperatures: Ea 1 ln k o = ln 5 C + A R 98 K Ea 1 ln k o = + ln 75 C A R 348 K Subtracting ln k k ln k 75 C ln k 5 C = E a R 1 348 K 1 98 K = ln k 75 C k 5 C ln 3000 M 1 s 1 80 M 1 s 1 = E a 1 8.314 J/mol K 348 K 1 98 K E a = 6.5 10 4 J/mol, or 6.5 kj/mol (c) We can use the rate 10 6 M and [O 3 ] = 5 10 9 M. We are given in the statement of the problem that k M 1 s 1. Rate = 80 M 1 s 1 (3 10 6 M) (5 10 9 M) = 1. 10 1 M/s (d) To use the Arrhenius equation to calculate the values of k, we have to first determine the value of the frequency factor A. To do this we can use the value for E a and k 4 1 1 6.5 10 J/mol 1 ln ( 80 M s ) = + ln A 8.314 J/mol K 98 K ln A = 9.608 1 A = 7. 10 So the value of k 4 6.5 10 J/mol 1 1 ln k = + ln(7. 10 ) 8.314 J/mol K 83 K ln k = 3.045 1 1 k = 1 M s The value of k 4 6.5 10 J/mol 1 1 ln k = + ln(7. 10 ) 8.314 J/mol K 308 K ln k = 5.0 1 1 k = 1.8 10 M s The values of k respectively, than the value of k k be used to determine the activation energy and the frequency factor values in this problem. 13.136. Collect and Organize Given the balanced reaction between NH 3 and HNO, we are asked to write the rate law and determine the units of the rate constant, k; compare the rate law to an alternate epression that includes the species NH 4 + and NO ; calculate the enthalpy of the overall reaction; and draw the reaction energy profile for the reaction occurring by a two-step mechanism.

Chemical Kinetics 105 This problem puts together many of the concepts in this chapter (writing rate laws from eperimental results, drawing a reaction profile for a multistep reaction) along with calculation of the enthalpy of a reaction from Chapter 5. (a) Because the reaction is first order in [NH 3 ] and second order in [HNO ], the rate law is Rate = k[nh 3 ][HNO ]. If the rate is in units of molar per second, then the units of k are M/s 1 M s ( M)( M ) = (b) Yes, the epression is equivalent to the rate law in part a. If the formation of NH + 4 and NO is an equilibrium step, then NH 3 + HNO = NH + 4 + NO k 1 [NH 3 ][HNO ] = k 1 [NH + 4 ][NO ] Solving for [NH 3 ][HNO ] gives k [ ][ ] 1 + k1 = NH3 HNO = NH 4 NO k 1 Substituting into the rate equation in part a gives (c) ΔH rn (d) = 1 mol N 0.0 kj/mol 1 mol NH 3 46.1 kj/mol = 48.4 kj Rate = k k 1 + NH k 4 NO HNO 1 + = k NH 4 NO HNO ( ) + ( mol H O 85.8 kj/mol) ( ) + ( 1 mol HNO 43.1 kj/mol) From the reaction energy diagram, we see that the second step is the rate-limiting step. 13.137. Collect and Organize We consider the mechanism for the echange of H 16 O around a Na + cation for H 18 O to write the rate law for the rate-determining step. We also need to think about the relative energies of the reactants and products if we are to sketch the reaction energy profile. (a) We are given that the first step of the reaction is rate determining, so this is the step from which we write the rate law epression. (b) In deciding which has the higher energy, the reactants or products, for the reaction profile, we need to consider the relative strength of the Na + 16 OH interaction versus that of the Na + 18 OH interaction.

106 Chapter 13 (a) Rate = k[na(h O) 6 + ] (b) Neither. The ion dipole interaction should not be significantly different for H 18 O versus H 16 O, so the energy of the reactants and the products in the reaction profile will be about the same. In reality, an isotope effect eists, in which the H 18 O Na + interaction is slightly stronger than the H 16 O Na + interaction, so the energy of the product in this reaction is slightly lower than the energy of the reactants. 13.138. Collect and Organize For the second-order decomposition of PAN, we are to calculate k for the reaction given t 1/, determine the rate of the reaction when the partial pressure of PAN is 10.5 torr, and draw a plot of P PAN versus time from 0 to 00 hr. This problem puts together many of the concepts in this chapter (using t 1/ to calculate k, calculating the rate of a reaction from the rate law, and plotting change in the concentration, here partial pressure, of a reactant over time). (a) From t 1/ we calculate k for this second-order reaction, where P PAN = 10.5 torr: 1 1 1 t 1/ =, or k = = k A 0 t 1/ A 100 hr 10.5 torr = 9.5 10 4 torr 1 hr 1 0 (b) For a second-order reaction, the rate law is (c) ( ) Rate = k A 0 = k PPAN = 9.5 10 4 torr hr ( 10.5 torr) = 0.105 torr/hr From the plot in part c, we may determine the instantaneous rate of the reaction at any time by determining the slope of the tangent line at that point. 13.139. Collect and Organize For the reaction of NO with ONOO, we are to use the provided data to determine the rate law for the reaction. We are also to draw the Lewis structure with resonance forms to describe the bonding in the ONOO anions and then use bond energies to estimate H rn.. (a) To determine the rate law, we can compare the effects of changing the concentrations of NO and ONOO on the rate of the reaction.

Chemical Kinetics 107 (b) After drawing the resonance forms for ONOO we can determine which is the preferred structure by assigning formal charges to the atoms in each resonance form. (c) The H rn may be estimated from bond energies by using = ΔH bond breaking + ΔH rn ΔH bond forming (a) For the order of reaction with respect to ONOO, we can compare eperiments 1 and : rate [ONOO ] 0, = rate 1 [ONOO ] 0,1 11 4 1.0 10 M/s 0.65 10 M = 11 4.03 10 M/s 1.5 10 M 0.50 = 0.500 = 1 The data table does not have two eperiments for which the concentration of ONOO stays the same. For the order of reaction with respect to NO, we can compare eperiments and 3 as long as we take into account the knowledge that the reaction is first order in ONOO. Between these two eperiments we see that as the ONOO concentration is quadrupled, we would epect that the rate would be quadrupled. The rate of the reaction when the NO concentration is halved simultaneously with the quadrupling of the rate on changing the ONOO, however, doubles: 11 rate3.03 10 M/s = 1.99 11 = rate 1.0 10 M/s Because the [ONOO ] = 4, which would quadruple the rate of the reaction, the [NO] must affect the rate by 1.99/4 = 0.498. Therefore, 4 0.65 10 M 0.498 = 4 1.5 10 M 0.498 = 0.500 = 1 The rate law for this reaction is Rate = k[no][onoo ]. We can use any eperiment to calculate the value of k. Using the data from eperiment 1:.03 10 11 M/s = k (1.5 10 4 M) (1.5 10 4 M) k = 1.30 10 3 M 1 s 1 (b) (c) H rn Bonds broken = O O (146 kj/mol) Bonds formed = N O (01 kj/mol) = [(1 mol 146 kj/mol) + (1 mol 01 kj/mol)] = 55 kj To solve this problem you had to draw on several concepts you have learned so far in this course. You had to review not only how to write a rate law given kinetic data but also how to draw resonance structures and how bond energies might be used to estimate the enthalpy of a reaction.

108 Chapter 13 13.140. Collect and Organize For the first-order decomposition of C 6 H 13 SNO, we can use a plot of ln[c 6 H 13 SNO] versus time to calculate the value of k. We are also asked which amino acids might be a source of S-nitrosothiols. The slope of the line of the plot of ln[c 6 H 13 SNO] versus time is equal to k. The amino acids that might be precursors to S-nitrosothiols must contain sulfur. (a) -6.8 ln[c 6 H 13 SNO] (ln M) -6.9-7 -7.1-7. -7.3 y = 6.50E 03 6.86E+00 R² = 1.00E+00 0 10 0 30 40 50 60 70 Time (min) k = slope = 6.50 10 3 min 1 S-Nitrosothiol drugs that generate NO in the body are being developed as treatments for respiratory and pulmonary diseases. 13.141. Collect and Organize Using data for [HNO ] over time, we can determine the order of the isotopic echange reaction with respect to [HNO ]. We are also asked whether the rate of the reaction will depend on [H 18 O]. We are given a single data set. If the plot of [HNO ] versus time yields a straight line, the reaction is zero order. If the plot of ln[hno ] versus time yields a straight line, the reaction is first order. If the plot of 1/[HNO ] versus time yields a straight line, the reaction is second order. (a) [HN 18 O ] 0.06 0.04 0.0 0.00-0.0 y = 0.0008 + 0.0384 R² = 0.61471 0 10 0 30 40 50 60 70 Time (min) Zero-Order Plot ln [HNO 18 O ] 0-1 - -3-4 -5-6 -7-8 -9 y = 0.073 3.8486 R² = 0.78631 First-Order Plot 0 10 0 30 40 50 60 70 Time (min)

Chemical Kinetics 109 1/[HN 18 O ] 500.00 000.00 1500.00 1000.00 y = 31.79 + 4.884 R² = 0.99993 Second-Order Plot 500.00 0.00 0 10 0 30 40 50 60 70 Time (min) This reaction is second order in [HNO ]. (b) Because the reaction miture has a very large [H 18 O], we cannot observe a change in [H 18 O] over time. The rate might be dependent on [H 18 O], but we cannot tell from the information given. If it were possible, we could place the reaction in a nonreactive solvent and then vary the [H 18 O] over time to determine the rate s dependence on its concentration. 13.14. Collect and Organize In the reaction of C H 4 with O 3, the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. Also, given the dependence of k on temperature, we are to calculate E a for this reaction. (a) To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two eperiments in which the concentration of that reactant changes but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law epression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the eperiments. Finally, we can calculate the rate of the reaction for two different concentrations of the reactants by substituting those concentrations and the calculated value of k into the rate law epression. (b) The Arrhenius equation is Ea 1 ln k = + ln A R T If we plot ln k (y-ais) versus 1/T (-ais), we can obtain a straight line with slope m = E a /R. The activation energy is therefore calculated from E a = slope R where R = 8.314 J/mol K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or e b = A. Once E a and A are known, we may use another form of the Arrhenius equation to calculate k at any temperature: k Aep E a = RT (a) Eperiments 1 and show that the rate dependence on [O 3 ] is 1: rate 1 = [O ] 3 1 rate [O 3 ]

110 Chapter 13 0.0877 M /s 0.0439 M /s = 0.86 10 M 0.43 10 M.00 =.0 = 1 Using this result, we compare eperiments and 3 to obtain the order of the reaction with respect to C H 4 : 1 The rate law for this reaction is Rate = k[o 3 ][C H 4 ]. Using eperiment 1 to calculate the value of k, (b) 0.43 10 M 1.00 10 M 0.0439 M/s = 0. 10 M 0.50 10 M 0.0110 M/s k = = 1 1.95 3.99 = 1 0.0877 M /s 0.0086 M ( ) ( 0.0100 M ) = 100 M 1 s 1 E a = slope R = 69.8 8.314 J/mol K =.19 10 4 J/mol, or 1.9 kj/mol Knowing the value of the activation energy, we can use the value of k at any of the temperatures given in the data table to calculate the value of the frequency factor, A. 13.143. Collect and Organize Using the raw data obtained for four eperiments in which the concentrations of NH and NO were varied, we are to write the rate law and determine the value of k for the reaction between NH and NO at 100 K. To determine the rate law, we can compare the effects of changing the concentrations of NH and NO on the rate of the reaction. Once we have determined the order of the reaction with respect to each reactant, we can write the rate law epression and use any of the eperiments to calculate the value of the rate constant, k. (a) For the order of the reaction with respect to NH, we can use the data from eperiments 1 and : rate = [NH ] 0, rate 1 [NH ] 0,1

Chemical Kinetics 111 0.4 M/s 0.1 M/s =.00 10 5 M 1.00 10 5 M.0 =.00 = 1 For the order of the reaction with respect to NO, we can use the data from eperiments and 3: rate [NO] 3 0,3 = rate [NO] 0, 5 0.36 M/s 1.50 10 M = 5 0.4 M/s 1.00 10 M 1.5 = 1.50 = 1 The rate law epression is Rate = k[nh ][NO]. (b) Using eperiment 1 to calculate the value of k, 0.1 M/s = k (1.00 10 5 M) (1.00 10 5 M) k = 1. 10 9 M 1 s 1 Any of the eperiments listed in the data table would give us the same value of k in the calculation in part b.